Answer:
(i have corrected the answers to 3 significant figures)
When dealing with projectile motion, we should consider its vertical and horizontal components.
Vertical velocity = 50sin60° m/s
Horizontal velocity = 50cos60° m/s
a) when it reaches the peak, meaning it can no longer travel further upwards, indicating the final vertical velocity is 0.
Take g=9.80665
using formula a = (v-u) /t
9.80665= (0-50sin60°) / t
t= 4.42 s
b) Consider the vertical component.
using formula t = 2u / g
t = 2(50sin60°) / 9.80665
t = 8.83s
c) Consider the vertical component again,
using formula H = u² / 2g
H = (50sin60°)² / 2(9.80665)
H = 95.6m
d) This is also the range of the projectile motion.
Using formula R = u²sin2θ / g
This time u should just be the initial velocity (neither horizontal/vertical)
R = 50²sin2(60) / 9.80665
= 221 m
- you can also do this by using s = ut formula, using the time calculated from b), but consider the u as the horizontal component.