Question

2) A football player kicked a ball at an angle of 37o above the horizontal with an initial velocity of 20 m/s, what is the maximum height reached by the ball?

a. 62 m
b. 26 m
c. 7.4 m
d. 5.1 m

Answer 1

Given:

A football player kicked a ball at an angle of 37° above the horizontal with an initial velocity of 20 m/s.

To find:

Max height reached by ball ?

Calculation:

The general expression of max height reached by a projectile is :

$H = \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

• Here 'u' is initial velocity and $\theta$ is angle of projection.

$\implies H = \dfrac{ {u}^{2} { \sin}^{2}( {37}^{ \circ} ) }{2g}$

$\implies H = \dfrac{ {20}^{2} \times { \sin}^{2}( {37}^{ \circ} ) }{2 \times 9.8}$

$\implies H = \dfrac{ {20}^{2} \times {( \dfrac{3}{5} )}^{2} }{19.6}$

$\implies H = \dfrac{ 400 \times \dfrac{9}{25} }{19.6}$

$\implies H \approx \: 7.4 \: m$

So, max height reached is approximately 7.4 metres (Option c)

Answer 2

Intial Velocity (u) = 20 m/s

Angle (θ) = 37°

----------------------------------------

$\boxed{ \rm H = \frac{ {u}^{2} \sin ^{2} ( \theta) }{2g} }$

$\sf➱ \: \: \frac{ {(20)}^{2} . \sin^{2}(37 \degree) }{2 \times 9.8}$

$\sf➱ \: \: \frac{400 \times ( \frac{3}{5})^{2} }{19.6}$

$\sf➱ \: \: \frac{400 \times \frac{9}{25} }{19.8}$

$\sf➱ \: \: 7.4m \: (approx.)$