Question

A football player kicks a ball at angle 35° to the horizontal with an initial velocity 15 ms. assuming that the ball travels in vertical plane calculate a) maximum height reached b) time of flight and c) horizontal range
[Given sin 35°=0.5735, sin 70°=0.9396​

Answer 1

Given: A football player kicks a ball at angle 35° to the horizontal with initial velocity 15m/s.

To find: 1- Maximum height reached

2- Time of flight

3- Horizontal range

Explanation: Initial velocity= 15 m/s

Angle= 35°

g= 10m/s^2

$height = \frac{ {u}^{2} { \sin( \alpha ) }^{2} }{2g}$

= (15)^2 * (0.5735)^2/2*10

= 225* 0.3289/20

= 3.7 m

$time = \frac{2u \sin( \alpha ) }{g}$

= 2*15*0.5735/10

= 1.7205 seconds

$range = \frac{ {u}^{2} \sin(2 \alpha ) }{g}$

= 15^2 sin (2*35)/10

= 225 * 0.9396/10

= 21.141 m

Therefore, the maximum height reached is 3.7 m, time of flight is 1.7205 seconds and horizontal range of projectile is 21.141 m.