Question

A boy standing on the top of a 40m high building projects a stone vertically upwards with an initial velocity of 10m/s.The stone eventually falls to the ground.(a)After how long will the stone strike the ground?(b)After how long will the stone pass through the point from where it was projected? (c) With what velocity will it strike the ground? (g=10m/s2)

Answer 1

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Answer 2

Answer: The time when the stone strike the ground is 4.05 s, The time when the stone pass through the point from where it was projected is 2.04 s and The velocity of the stone when it will strike the ground is 29.73 m/s.

Explanation:

Given that :

Height h = 40 m

Initial velocity u = 10 m/s

(a). The time when the stone strike the ground is

Using equation of motion

$s= ut-\dfrac{1}{2}gt^2$

$-40=10t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-10t-40=0$

$t= 4.05 s$

(b). The time when the stone pass through the point from where it was projected is

Using equation of motion

$0=10t-4.9t^{2}$

$4.9t^{2}=10t$

$t = \dfrac{100}{49}$

$t=2.04 s$

(c). The velocity of the stone when it will strike the ground is

Using equation of motion

$v^2-u^2=2as$

$v^2=100+2\times9.8\times40$

$v= 29.73 m/s$

Hence, The time when the stone strike the ground is 4.05 s, The time when the stone pass through the point from where it was projected is 2.04 s and The velocity of the stone when it will strike the ground is 29.73 m/s.