Question

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

Answer 1

option B

Step-by-step 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 = (1)/(2)gt^2`

`t = \sqrt{(2s)/(g)}`

`t = \sqrt{(2* 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 =
`(65)/(4.53)` =14.35 m/s

speed of the ball =
`√(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