Question

A ball thrown vertically upward rises to a height of 44.1m. calculate the initial velocity of the ball.

Answer 1

u=?;v=0;a=-9.8m/s^2;s=44.1mts
Using third equation of motion => v^2=U^2+2*a*s
substituting we get 0^2=u^2+2*-9.8*44.1=>u^2=864.36
=>u=29.4m/s

Answer 2

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

$\green{\therefore{\text{Initial\:velocity=29.7\:m/s}}}$

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

• In the given question information given about a ball is thrown vertically upward rises to a height of 44.1 m.

• We have to calculate its initial velocity.

$\green{ \underline \bold{Given : }} \\ : \implies \text{Height(s) = 44.1\: m} \\ \\: \implies \text{Acceleration(a) = - 10 \: {m}/s}^{2} \\ \\ : \implies \text{Final \: velocity(v) = 0 \: m/s} \\ \\ \red{ \underline \bold{to \: find : }} \\ : \implies \text{Initial \: velocity = ?}$

• According to given question :

$\bold{Using \: third \: equation \: of \: motion} \\ : \implies {v}^{2} = {u}^{2} + 2as \\ \\ : \implies 0 = {u}^{2} + 2( - 10) \times 44.1 \\ \\ : \implies {u}^{2} =882 \\ \\ : \implies {u} = \sqrt{882} \\ \\ \green{: \implies \text{{u} = 29.7\: ms}} \\ \\ \green{\therefore \text{Initial \: velocity \: of \: ball \: is \: 29.7 \: m/s}}$