Question

A football is kicked towards a goal keeper with an initial speed of 20 m/s at an angle of 450 with the horizontal. At the moment the ball is kicked, the goal keeper is 50 m from the player. At what speed and in what direction must the goal keeper run in order to catch the ball at the same height at which it was kicked?

Answer 1

Given that,

Initial speed = 20 m/s

Angle = 45°

Distance s= 50 m

We need to calculate the range

Using formula of range

$R=\dfrac{u^2\sin2\theta}{g}$

Where, u = speed

Put the value into the formula

$R= \dfrac{20^2\times\sin90}{9.8}$

$R=40.81\ m$

We need to calculate the time

Using formula of time of flight

$T=\dfrac{2u\sin\theta}{g}$

Put the value into the formula

$T=\dfrac{2\times20\sin45}{9.8}$

$T=2.8\ sec$

We need to calculate the distance of the end of projectile from the receiver

Using formula of distance

$D=R-s$

Put the value into the formula

$D=40.81-50$

$D=-9.19\ m$

Negative sign shows the direction of goalkeeper

We need to calculate the speed of the goalkeeper

Using formula of speed

$v =\dfrac{D}{T}$

Put the value into the formula

$v=\dfrac{9.19}{2.8}$

$v=3.28\ m/s$

Hence, The speed of goalkeeper is 3.28 m/s and the direction of goalkeeper toward the player.