Q#27
A healthy young man standing at a distance of 7 m from an 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height at (1.8 m)?Answer:
Height of fall = h = 11.8 – 1.8 = 10 m,
Let time of fall be t. From the equation
h = ut + ½gt² = ½gt²
t = √(2h/g) = √(20/9.8)
Now the young man has this time to run 7 m to catch the kid. So, his speed should be
= 7/√(20/9.8) = 7/1.428 = 4.9 m/s
Q#28
An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant, the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform.
Answer:
h = 12.1 m, As in question 27 time of fall,
t = √(2h/g) = √(2 x 12.1/9.8) = 1.57 s
Speed of NCC parade = 6 km/h = 6000/3600 = 5/3 m/s
So, the distance of the cadet who will receive the berry on his uniform from the tree
= (5/3)(1.57) = 2.62 s
Q#29
A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g = 10 m/s².
Answer:
Time of fall t = √(2h/g).
Time of fall for h – 6 m = √{2(h – 6)/g}
Difference of time = √(2h/g) – √{2(h – 6)/g} = 0.2, we solve it for h.
√(2h) – √{2(h – 6)} = 0.2√g squaring both sides we get,
2h + 2(h – 6) – 2√(4h² – 24h) = 0.04g
4h – 12 – 0.04g = 2√(4h² – 24h)
2h – 6.2 = √(4h² – 24h), Again squaring both sides we get,
4h² – 2(2h)(6.2) + (6.2)² = 4h² – 24h
24.8 h – 24h = 6.2²
h = (6.2)(6.2)/0.8 = 48.0 m
Q#30
A ball is dropped from a height of 5.0 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.
Answer:
Velocity just near the floor v is given by,
v² = u² + 2gh, here u = 0, h = 5 m,
v² = 2g(5) =10g
This v is the initial velocity u for the penetration in sand and final velocity v = 0, here h = 10 cm = 0.10 m, Let r be the retardation of the ball in sand. From the equation v² = u² - 2ah, we have,
0 = 10g – 2r(0.10)
r = 10g/0.2 = 50g = 490 m/s²
Q#31
An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.
Answer:
Let the acceleration of the elevator be a'. Since the coin is falling freely, its acceleration is due to the gravity, g. As viewed from the frame of the elevator the observer sees the relative acceleration of the coin due to the change of frame. This relative acceleration is = g – a'.
Now using the equation
H = ut + ½at², we have u = 0, h = 6 ft, taking g = 32 ft/s², t = 1 s,
6 = ½(g – a')(1)²
g – a' = 12,
a' = g – 12 = 20 ft/s².
Q#32
A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground. (c) the velocity (direction and magnitude) with which it strikes the ground.
Answer:
(a) Let the time to reach the ground be t. Since the ball is thrown horizontally its downward component of velocity will be zero means u=0. h = 100 m. From the equation of accelerated motion h
= ut + ½gt² we get,
100 = ½(9.8)t²
t² =100/4.9
t = 4.5 s
(b) The horizontal movement will be a uniform speed. Distance traveled before reaching the ground = speed x time
= 20 m/s x 4.5 s = 90 m
(c) At the time it strikes the ground, the horizontal component of velocity will be Vx = 20 m/s
Vertical component of the velocity
Vy = u + gt = 0 + 9.8(4.5)
Vy = 44.1 m/s.
So, the magnitude of the velocity before it strikes the ground
= √{(Vx)² + (Vy)²} = √(20² + 44.1²) = 48.5 m/s ≈ 49 m/s
If θ be the direction of velocity with horizontal.
tanθ = Vy/Vx = 44.1/20 = 2.205
θ = 65.61 ≈ 66°
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