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nalin [4]
3 years ago
11

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

away from and below the point of release. What is the speed of the ball just before it strikes the ground?
a) 43 m/s
b) 47 m/s
c) 39 m/s
d) 36 m/s
e) 14 m/s
Physics
1 answer:
aivan3 [116]3 years ago
5 0

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m    

speed at which the ball strike = ?

vertical velocity  = 0 m/s

time at which the ball strike

s = \dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 100}{9.8}}

t = 4.53 s

vertical velocity at the time  4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = \dfrac{65}{4.53} =14.35 m/s

speed of the ball = \sqrt{44.39^2+14.35^2}

                             = 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

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