Question

with what speed must a ball be thrown vertically up in order to rise to a maximum height of 53.7m

Answer 1

Hey.

$as \: maximum \: height \: = 53.7 \: m \\ \\ the \: velocity \: at \: the \: maximum \: height \: \\ will \: be \: 0 \: m {s}^{ - 1} \\ \\ tha \: accleration \: \: a = g = - 9.8 {m}^{ - 2} \\ \\ let \: the \: velocity \: of \: throw \: be \: u \: m {s}^{ - 1} \\ \\ as \: \: {v}^{2} - {u}^{2} = 2as \\ \\ so \: \: \: {0}^{2} - {u}^{2} = 2( - 9.8)53.7 \\ \\ so \: \: {u }^{2} = 1052.52 \\ \\ so \: \: = \: 32.44 \: m {s}^{ - 1}$


Hence, if the ball is thrown at a speed of 32.44 m/s vertically upwards it will gain a vertical height of 53.7 m.

Thanks

Answer 2

Answer:

$32.44ms^{-1}$

Explanation:

By the 3rd equation of motion, $v^{2}-u^{2}=2as$

Given: $v=0ms^{-1}$

$u=?$

$s=53.7m$

$a=-9.8ms^{-2}$

Substituting;

$u^{2}=\sqrt{v^{2}-2as}$

$u=\sqrt{\left(0\right)^{2}-2\left(-9.8\right)\left(53.7\right)}$

$u=\sqrt{2\cdot9.8\cdot53.7}$

$u=\sqrt{1052.52}$

$u=32.44ms^{-1}$

Hope this helps you. Good evening.