Question

A ball is thrown vertically upward rises a height of 20 m . Calculate a) The velocity with which the ball is thrown up and b) the time taken by the ball to reach the heighest point. (g= 10m/s²)

Answer 1

Height covered by the ball = h = 20 m

Final velocity (At the heighest point) = 0 m/s

Acceleration = g = 10m/s² (Given in statement)

By the equation , v²-u² = 2as :

0²-u² = 2×(-10)×20

u²= 2×10×20

u² = 400

u= 20 m/s

Then by the equation :

v=u+at

0= 20+(-10)×t

t = 2s

Answer 2

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The answer goes here....

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Given,

h = 20 m

g = 10 m/s²

Final velocity = $0\:m{s}^{-1}$

Now, by using the equation -

⇒ ${v}^{2}-{u}^{2}$ = $2gh$

⇒ ${v}^{2}$ = ${u}^{2}+2gh$

⇒ $0$ = ${u}^{2}+2\times (-10)\times 20$

⇒ ${u}^{2}$ = $400$

⇒ $u$ = $20\:m{s}^{-1}$

So, the velocity by which the ball is thrown up is $20\:m{s}^{-1}$

Now, the time taken -

⇒ $v$ = $u+gt$

⇒ $0$ = $20+10t$

⇒ $t$ = $2\:sec$

So, the time taken by the ball to reach the highest point is 2 sec.

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Thanks !!..