Question

A soccer ball is kicked at an angle of 45° with a speed of 20 m/s. What is the max height? How Iong does it take to reach max height? What is the hang time?

Answer 1

`0.5` metres and `sqrt(2)` seconds

Explanation:

Maximum height `= (u^2sin^2theta) / (2g) = [(20 ms^-1)^2 × (1/sqrt(2))^2] / (2 × 10 ms^-2) = 0.5 m`

Time to reach maximum height `=(usintheta)/g = (20 ms^-1 × 1/sqrt(2)) / (10 ms^-2) = sqrt(2)` second

[Note: I took g = 10 m/s²]