Question

A ball is thrown vertically upward rises to a height of 45 m. Calculate Initial velocity of ball

Answer 1

Answer

The initial velocity of the ball is 29.7 m/s

Explanation:

Given :

The height attuned by the ball. h=45 m

At the maximum height of its motion the final velocity of the ball will be zero. Let u be the initial velocity

According to third equation of motion we have

$2gh=v^2-u^2$

where

• g is acceleration due to gravity $g=9.8\ m/s^2$
• h is the displacement
• v is the final velocity
• u is the initial velocity

$2\times 9.8\times 45=0^2-u^2\\u=29.7\ \rm m/s\\\u=29.7\ \rm m/s$

hence the initial velocity of the ball is calculated

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Initial\:velocity=30\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a ball is thrown vertically upward rises to a height of 45 m.

• We have to calculate its initial velocity.

$\green{ \underline \bold{Given : }} \\ : \implies \text{Height(s) = 45\: m} \\ \\: \implies \text{Acceleration(a) = - 10 \: {m}/s}^{2} \\ \\ : \implies \text{Final \: velocity(v) = 0 \: m/s} \\ \\ \red{ \underline \bold{to \: find : }} \\ : \implies \text{Initial \: velocity = ?}$

• According to given question :

$\bold{Using \: third \: equation \: of \: motion} \\ : \implies {v}^{2} = {u}^{2} + 2as \\ \\ : \implies 0 = {u}^{2} + 2( - 10) \times 45 \\ \\ : \implies {u}^{2} =900 \\ \\ : \implies {u} = \sqrt{900} \\ \\ \green{: \implies \text{{u} = 30 \: ms}} \\ \\ \green{\therefore \text{Initial \: velocity \: of \: ball \: is \: 30 \: m/s}}$