Question

A ball is thrown vertically upward with a velocity of 49m/s calculate max height reached by the ball

Answer 1

2gh=v^2 - u^2
2×9.8×h=0-2401
19.6 h=-2401
h=2401/19.6
h = 122.5 m

Answer 2

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

$\green{\therefore h_{max}=120.05\:m}$

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

• In the given question information given about a ball thrown vertically upward it means the ball is thrown at an angle of 90°.We have to find the maximum height.

$\green{\underline \bold{Given :}} \\ \\ :\implies \text{Initial \: velocity(u) = 49 \: m/s} \\ \\ :\implies \text{Final \: velocity(v) = 0} \\ \\ :\implies \text{Angle \: of \: projection = 90 \degree} \\ \\ \red{\underline \bold{To \: Find :}} \\ \\ :\implies h_{max} = ?$

• According to given question :

$\: \: \: \: \: \green{ \bold{ First \: method : }}\\ \\ \bold{By \: formula \: of \: h_{max} : } \\ :\implies h_{max} = \frac{ {u}^{2} { \sin^{2} \theta} }{2g} \\ \\ :\implies h_{max} = \frac{ {49}^{2} \times { \sin^{2} 90\degree} }{2 \times 10} \\ \\ :\implies h_{max} = \frac{ 2401\times {1}^{2} }{20} \\ \\ \bold{:\implies h_{max} = 120.05 \: m}$

$\: \: {\green{ \bold{Alternate \: method : } }}\\ \\ \bold{By \:second\:equation\: of \: motion : } \\ :\implies {v}^{2} = {u}^{2} + 2as \\ \\ :\implies {0}^{2} = {49}^{2} + 2 \times( - 10 )\times s \\ \\ :\implies - 900 = - 20 \times s \\ \\ :\implies s = \frac{ - 2401}{ - 20} \\ \\ \green{:\implies s = 120.05\: m}$