a football is kicked toward a goal keeper with an initial speed of 20m/s at an angle of 45 degrees with the horizontal .at the moment the ball is kicked the goal keeper is 50m from the player .at what speed and in what direction must the goalkeeper run in order to catch the ball at the same height at which it was kicked

3.18 m/s

Explanation:

Given that

Initial speed of the ball, u = 20 m/s

Angle of inclination, θ = 45°

Distance from the ball, h = 50 m

Using equations of projectile to solve this, we have

We start by finding the time of flight, T

T = 2Usinθ/g

T = (2 * 20 * sin45)/9.8

T = (40 * 0.7071) / 9.8

T = 28.284/9.8

T = 2.89 s

Next we find the Range, R

R = u²sin2θ/g

R = (20² * sin 90) / 9.8

R = (400 * 1) / 9.8

R = 400/9.8 = 40.82 m

Distance the gk must cover

40.82 - 50 m

-9.18 m or 9.18 m in the opposite direction.

Speed of the GK = d/t

9.18 / 2.89 = 3.18 m/s