Vectors and Projectile motion

Question

A cannonball is fired at 50m/s at an angle of 30°above the horizontal

A)What is the maximum height the cannonball reaches?

B)How far does the cannonball land?

Answer 1

The vertical component of velocity = v sin t = 50 sin 30 = 25 m/s

The horizontal component of velocity = v cos t = 50 cos 30 = 43.3 m/s

Time to reach max height vf = vu + at so 0 = 25 - 10t, so t = 2.5s (v = 0 at max height)

i) h = vut +0.5 at^2 = 25*2.5 - 0.5 * 10*2.5^2 = 31.25 m

ii) x = horizontal component of velocity * total time

total time = 2.5s * 2 = 5 s

x = 43.3 * 5 = 216.5 m