Projectile motion question (1 Viewer)

Exp

Member
Hey! Could you guys help me with this problem? You don't have to answer all of the parts, just help me get started A ball is thrown horizontally from the top of a cliff and hits the ground 6 seconds later at 30 degrees to the vertical, moving at 67.9 m/s. Find:
a) the initial velocity of the projectile
b) initial horizontal and vertical velocity
c) range
e) final horizontal and vertical velocity

Sp3ctre

Active Member
Some prep work I did before answering the questions:
Uy = 0

Vy = 67.9 cos 30
= 58.80... m/s
Vx = 67.9 sin 30
= 33.95 m/s

a) Uy = 0 (launched horizontally)
Ux = Vx
= 33.95 m/s

Therefore initial velocity = 33.95 m/s horizontally (at 0 degrees to the horizontal)

b) Initial horizontal velocity = 33.95 m/s
Initial vertical velocity = 0 m/s

c) Range = Ux*t
= 33.95 * 6
= 203.7 m

e) All worked out already in prep work again:
Final horizontal velocity = 33.95 m/s
Final vertical velocity = 58.80... m/s (to 2 d.p)

Hope that all makes sense. I am very prone to making careless mistakes too so let me know if one or more of my answers do not match your answers.

Exp

Member
Yep, your answers are spot on! Thank you so much for the help 