## Lakshya Education MCQs

Question: A person standing on the top of a cliff, 171 ft high, has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall? Give g = 32 ft/s2.
Options:
 A. 36 feet B. 12 feet C. 0 feet the packet reaches his friend D. 192 feet
: D In the given figure we can find θ by using trigonometry. tanθ = 171228 = 34 θ = 37. This means the ball was thrown at an angle 37 to horizontal also its given that it was thrown with velocity 15 ft/s. The dotted line shows how the trajectory of the ball thrown So we know 2 D is nothing but 2 one dimensional motion. Let's break the components: uy = 15sin37 = 15×35 = 9ft/s ay = 32ft/s2 sy = 171ft. sy = uyt+12ayt2 171 = 9t322t2 16t2+9t171=0 16t2+57t48t171=0 t = 3sec We need to know in that time the projectile covered how much of horizontal distance. So ux=15cos37 = 15×45 = 12ft/s ax = 0; t = 3 sx = uxt = 12×3 = 36ft The packet went only 36 ft towards his friend The packet fell short by (22836) = 192ft.

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## More Questions on This Topic :

Question 1. 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 successful shot. Assume that the point of Projection and the edge of the boat are in the same horizontal level.
1.    15∘,75∘
2.    18.5∘,75∘
3.    15∘,71.5∘
4.    18.5∘,71.5∘
: A Let AB be the boat, to touch A, Range should be 5m.
u= 10 ms, g = 10m/s2, R = 5 m,
R = u2sin2θg 5 = 102sin2θ10 sin 2θ = 12
2θ = 30,15θ(as inπ-θ=sinθ)
θ = 15,75
To throw at B
R = 6m
6 = 102sin2θ10sin 2θ = 35
2θ = 37,143(sin(π-θ)= sin θ)
θ = 18.5,71.5
Minimum is 15 and maximum is 75
For a successful shot,
15θ18.5 and 71.5θ75
Minimum is 15 and maximum is 75. .
Question 2. The figure shows two positions A and B at the same height h above the ground. If the maximum height of the projectile is H, then determine the time t elapsed between the positions A and B in terms of H.

1.    t = √8hg
2.    t = √h8g
3.    t = √8g(H−h)
4.    t = √(H−h)8g
: C Time of flight for this part of motion (between points A and B, both at height h) assuming speed of projection to be v and angle of projection to be α T=2vsinαg ---------(I) Now we have to eliminate v & α as these variables we assumed. Hmmmmm..........How do we do that, okay what else is given? Yes, the maximum height i.e. (Hh). (Hh) = v2sin2α2g ----------(II) Thank goodness! Now from equation (I) vsinα = Tg2 Equation (II) (Hh) = T2g24(2g) (Hh) = gT28 T = 8(Hh)g
Question 3. An aeroplane moving horizontally with a speed of 720 km/h drops a food packet, while flying at a height of 396.9 m. The time taken by a food packet to reach the ground and its horizontal range is (Take g=9.8 m/sec2)
1.    3 sec and 2000 m
2.    5 sec and 500 m
3.    8 sec and 1500 m
4.    9 sec and 1800 m 