47

Q 47. A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. 

(a) If he heads in a direction making an angle q with the flow, find the time he takes to cross the river.

(b) Find the shortest possible time to cross the river.

Answer: (a) The component of velocity perpendicular to direction of flow of river = 3sinθ  km/h =3000sinθ/60 m/min = 50 sin θ  m/min. 

Time taken to cross the river = 500 m/ 50sin θ  min = 10/sin θ  min. 

(b) For the shortest possible time the denominator "sin θ" should be maximum. And maximum possible value of "sin θ" is 1 for θ=90°. So shortest possible time = 10/1  min =10 minutes.

46

 A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. 

(a) Find the time taken by the boat to reach the opposite bank. 


(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank? 

45

A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Sol. When the apple just touches the end B of the boat. 
x = 5 m, u = 10 m/s, g = 10 m/s2, θ = ? 
x = (u^2 sin⁡〖2 θ〗)/10 
⇒ 5 = (〖10〗^2 sin⁡〖2 θ〗)/10⇒ 5 = 10 sin 2 θ 
⇒ sin 2 θ = 1/2 ⇒ sin 30° or sin 150° 
⇒ θ = 15° or 75° 

Similarly for end C, x = 6 m 
Then 2 θ1 = sin–1 (gx/u2) = sin–1 (0.6) = 182° or 71°. 
So, for a successful shot, θ may very from 15° to 18° or 71° to 75°. 

44

The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane

of motion and the first bench is 110 m from the batsman. On which bench will the ball hit?

One way of doing this is to find an equation for the trajectory of the ball. We'll also find an equation for the benches, 

I'll interpret the width of the benches as being the distance from the front to back. For the purpose of this question we can take the benches to be a straight line that goes up 1 m for every 1 m back. You know how a stadium is right, the seats further back are higher. It's not true that the seats at the back are at the same height at the seats in front, otherwise the fans in the back will see nothing! 

So, we know is line will have a gradient of 1. If we take our origin to be the point where the ball was hit.then the benches will follow the line y = x - 110. 

Now we use the kinematic equations of motion to find an equation for the (parabolic) trajectory the ball traces out. We know it will be an equation of the form y = a(x-h)² + k, where k is the height of the highest point and h is the horizontal coordinate of that point. We also know, from the kinematic equation in the vertical direction d = ut + 1/2 gt², that a = g/2 = -9.8/2 = -4.9. 

So the parabola is y = -4.9(x-h)² + k. 

Let's find the values of k and h now. 
k first: We use the kinematic equations of motion. We know this is the height of the highest point, and at this point, the vertical component of the speed is 0 m/s since it will have stopped going up but not yet begun coming down. 

To find k: 
u = initial velocity = 35sin53 = 27.95 m/s

v = final velocity = 0 m/s 

k = vertical distance = ? 
a = acceleration due to gravity = -9.8 m/s² 

v² = u² + 2ak, so since v² = 0, k = -u²/2a, and plugging in the numbers gives k = 39.86 m. 

To find t, the time parameter which will help find h: 
v = u + at, so t = -u/a = 2.85 s. 

Now we turn our attention to the horizontal direction to find h. This will be the horizontal coordinate at time 2.85 s. We know that the only force acting on the ball is the purely vertical force of gravity, so there are no horizontal forces acting on the ball, therefore no horizontal acceleration and therefore the horizontal component of velocity is constant. And the horizontal component of velocity is 35cos53. So horizontal distance = horizontal velocity times time = (2.85 s)(35cos53 m/s) = 60.07 m. 

So the trajectory of the ball is given by 

y = -4.9(x - 60.07)² + 39.86, which can be written 
y = -4.9(x² - 120.14x + 3609) + 39.86, or 
y = -4.9x² + 589x -17644 

The point where the ball hits the benches will be where this line intersects with the line y = x - 110. 

So we solve -4.9x² + 589x - 17644 = -x - 110, or 
-4.9x² + 588 x + 17754 = 0, a quadratic equation with solutions obtained using the quadratic formula of 145 and a negative solution we may disregard. 

So the y-coord is y = x - 110 = 145 - 110 = 35 m high, so it hits the 35th bench.

43

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.

42

Q 42. A staircase contains three steps each 10 cm high and 20 cm wide (figure). What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane?

Answer: To hit directly the lowest plane the ball will have to clear point B. From A to B the ball will move a horizontal distance equal to 40 cm = 0.4 m and the vertical distance equal to 20 cm =0.2 m. Let t be the time taken to reach point be. In this time vertical movement under the gravity has following data, 

u=0, h=0.2 m, from the equation h=ut+½gt² , we get,

0.2 = 0+½*9.8 t²    → t²=0.4/9.8  = 0.04  → t=0.2 

Let v be the horizontal velocity of the ball, then the horizontal distance traveled by the ball = 0.2 v . In order to clear point B this must be at least =0.4, 

0.2 v= 0.4  → v=2 m/s. So minimum velocity required to to hit the lowest floor  is 2 m/s.  

Let h be the height and w be the width of each step.

Let O be the uppermost plane and A be the lowermost plane.

 

 

 

 

Let the ball roll from the uppermost plane O with a horizontal velocity 'v' .Here the vertical velocity (velocity in the y direction will be 0)

If 't' is the time taken by the ball to reach the lowest plane A and 'x' the horizontal range, then we have:

x = v t  

t = x /v        ---------------------(1)

 

Here the horizontal range of the ball lies between 2w and 3 w, or we can say that the horizontal range (x) of the ball will be greater than 2w.

Given h = 10 cm =0.10 m,

         w = 20 cm =0.20 m

i.e. x ≥ 2w

x ≥ 2×0.20m

x≥ 0.40 m

 

The vertical height, through which the ball travels from the plane O to plane A = 2h

From the equation of motion, s=ut+(1/2) gt² 

Here u =o for vertical motion so:

2h = (1/2)gt² 

 

From (1) and (2)  we get that

 

So the minimum horizontal velocity of a ball rolling off from the uppermost plane(O) so as to hit directly the lowest plane(A) of the staircase = 2 m/s. Further for the ball to roll from the point O to reach the ground level the horizontal velocity should be greater than 2 m/s.

TEST

1. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1m/s^2 and the projection velocity in the vertical direction is 9.8m/s. How far behind the boy will the ball fall on the car? (Ans:2m)
2. A person standing on top of a cliff 171ft high has to throw a packet to his friend standing on the ground 228ft horizontally away. If he throws the packet directly aiming at his friend, with a speed of 15ft/s, how short will the packet fall? (Ans:192ft)
3. A shell is fired horizontally with a velocity of 200m/s from the top of a tower. It hits a target on the ground after 2s. (i) what is the height of the tower? (ii)Where is the target situated? (iii) with what vertical velocity does the shell strike the target? (Ans:19.6m, 400m, 19.6m/s)
4. From the top of a building 19.6m high, a ball is projected horizontally. After how long will the ball strike the ground? If the line joining the point of projection to the point where it strikes the ground is 60degree with the horizontal, what is the velocity with which the ball is projected? (Ans: 2s, 5.65m/s)
5. A ball is projected horizontally with a velocity of 5m/s. Find the position and velocity after 0.25s. (Ans: 1.29m, 5.57m/s)
6. A man can swim at a speed of 3km/h in still water. He wants to croos a 500m wide river flowing at 2km/h. He keeps himself always at an angle of 120degrees with the river flow while swimming. Find (a)the time he takes to cross the river. (b) at what point on the opposite bank will he arrive? (Ans: 1/3 sq.root3 h , 1/6 sq.root3km)