Question

Saad throws a ball at an initial velocity of 42 m/s Calculate the maximum distance the ball can reach assuming the ball is caught at the same height at which it was released

Answer 1

Answer:

for maximum height we have formula

u^2/2g =42×42/2×10=88.2 m

Answer 2

Answer:180m

Explanation: its quite easy question is about max distance so it means max range for that we have formula

Rmax=vi^2/g

Value of vi is=42m/s and g=9.8 so by inserting values we have

R max =(42)^2 /9.8 =1764/9.8 =180 m

So ans is 180 m thanks