Show all work and answer all 4 parts. Projectile motion. 20 points. Thank you.

(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.

it may be wrong but i think its a