u = 60m/s, θ = 60°
1. Maximum height
H = \(\frac {u^2 sin^2 (\theta)}{2g}\)
= \(\frac {60^2 sin^2 (60)}{2 \times \,9.8}\)
= 137.74 m.
2. Time to reach maximum height
t = \(\frac {u sin (\theta)}{g}\)
= \(\frac {60\times sin (60)}{9.8}\)
= = 5.30s.
3. Range R = \(\frac {u^2 sin (2\theta)}{g}\)
= \(\frac {60^2 sin(2 \times 60)}{9.8}\)
= 318.12 m.