Answer:
Minimum average speed with which the opponent must move to hit the ball is 5.56 m/s.
Step-by-step explanation:
To find the average speed of the opponent, we must find the distance he has to cover and the time he has to cover it. For this we need to find:
a. time at which the ball is at 2.10 m above the launch point
b. distance of ball from the launch point at that time.
We know that
initial velocity = Vi = 15 m/s
Horizontal component of Vi = Vx = Vi*cos(50°) = 9.64 m/s
Vertical Component of Vi = Vy = Vi*sin(50°) = 11.50 m/s
a. Using 2nd equation of motion, we will find time 't' taken by the ball to reach the height 2.10 m
h = Vy*t - 0.5*g*t²
2.1 = 11.5t - 0.5(9.8)t²
4.9t² - 11.5t + 2.1 = 0
Implies that t = 2.15 s or t = 0.20 s
We know that the ball will reach 2.1 m height twice, first while going up(t = 0.20s) and second while coming down(t = 2.15s). Opponent begins moving after time 'T' of hitting the ball where T = 0.22s. As T > (t = 0.20s), it means the opponent will reach the ball when it is coming down and t at that point will be equal to 2.15s.
b. Finding distance 'x' of ball from launch at t = 2.15s:
x = Vx * t
x = 9.64*2.15
x = 20.73 m
Finding the distance 'd' opponent needs to cover to reach the ball at height 2.10 m:
d = (distance of ball at height 2.10 m) - (distance between the two players)
d = 20.73 - 10
d = 10.73 m
Finding time 't°' opponent has to reach the ball:
t° = Time taken by the ball to reach the hitting point - Time player took to decide his move
t° = 2.15 - 0.22s
t° = 1.93 s
Average Speed of the opponent = d/t°
= 10.73/1.93
= 5.56 m/s