Q#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.Answer:
Time of flight of the ball = 58.8/14.7 s = 4 s
(a) Since the ball returns back to the truck its angle of projection is vertically upwards with respect to the truck. Let u be the speed of projection, we have t=4 s, displacement h=0 m, using equation
h = ut – ½gt²
0 = 4u – ½(9.8)16
u = 9.8(8/4) = 19.6 m/s
(b) As seen from the road the speed of the ball will be resultant of two speeds, vertical speed u = 19.6 m/s and horizontal speed given by the truck v = 14.7 m/s,
we get it = √(u² + v²) = √(19.6² + 14.7²)
= 24.5 m/s and the angle seen from the road will be
= tan-1(19.6/14.7) = tan-1 1.33 = tan-1 (4/3) = 53° with horizontal.
Q#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 m 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?
Answer:
Distance of first bench = 110 m, Let us assume that the ball strikes the nth bench. The height above the strike level = (n – 1) m and horizontal distance traveled = (110 + n) m
Now from the equation h = ut - ½gt²
(n – 1) = 35 sin 53° t – ½(9.8)t²
4.9t² - 28t + (n – 1) = 0
t = [28 ± √{28² - 4(4.9)(n – 1)}]/9.8 (we shall take the +ve sign because the ball will strike the bench when falling.
t = 28/9.8 + √{784 – 19.6(n – 1)}/9.8 = 2.86 +√{8.16 – 0.204(n – 1)}
Horizontal distance traveled = 35 cos 53° t = 110 + n
(21.06)[2.86 + √{8.16 – 0.204(n – 1)}] = 110 + n (Put value of t)
60.24 + 21.06√{8.16 – 0.204(n – 1)} = 110 + n
21.06√{8.16 – 0.204(n – 1)} = 49.76 + n, squaring both sides we get,
(444){8.16 – 0.204(n – 1)} = 2476 + 99.52n + n²
n² + 99.52n + 90.58n + 2476 – 3623 + 90.58 = 0
n² + 190.1 n – 1056.42 = 0, solving this quadratic equation for n and taking only the positive value we get n = 5.4
It means the ball clears the 5 th bench and hits the sixth bench.
Q#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 center 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 a successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.
Answer:
Since the length of the boat is 1.0 m and the man is sitting 5.5 m away from the center of the boat, for the apple to fall in the boat the range of projection should be minimum 5.0 m and maximum 6.0 m. We have u = 10 m/s.
Range x = (u² sin 2θ)/g Where θ is the angle of projection.
sin 2θ = gx/u² for x = 5.0 m
sin 2θ = 9.8(5/100) = 0.49
2θ = 30° or 150°
θ = 15° or 75°
For x = 6.0 m
sin 2θ = 9.8(6/100) = 0.59
2θ = 36° or 144°
θ = 18° or 72°
So, for the successful shot minimum angle of projection is 15° and maximum 75°. In fact, the successful shots will be between angles 15° to 18° and 72° to 75°. between 18° to 72° the shot will not be successful.
Q#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?
Answer:
(a) Velocity of the boat perpendicular to river u = 10 m/s. Width of the river = d = 400 m. Time taken by the boat to reach opposite bank = d/u = 400/10 s = 40 s.
(b) In this 40 s time, the boat will go downstream with a velocity of 2.0 m/s.
So, the boat will reach opposite bank a distance = 2 x 40 = 80m from the point directly opposite the starting point.
Q#47
A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to the water is 3 km/h. (a) if he heads in a direction making an angle θ 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 = 3 sinθ 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 the maximum possible value of "sin θ" is 1 for θ = 90°. So, shortest possible time = 10/1 min = 10 minutes.
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