Calculating Impact Speed of a Ball Thrown Toward a Cliff

[HaLAA]

Homework Statement

A ball is thrown toward a cliff of height h with a speed of 28.0 m/s and an angle of 60.0° above horizontal.
What is the ball's impact speed

Homework Equations

Vf=Vi-at
Sf=Si+Vit-1/2gt^2
Vy=sintheta
Vx=costheta

The Attempt at a Solution

first find the vertical velocity = 28 sin 60 = 24.25m/s

find time to maximum height from Vf = Vi+ at
0 = 24.25 - 9.8t
t = 2.47sec

the height reached
v^2 = u^2 + 2as
0 = 24.25^2 - 2 x 9.81 s
s = 29.97m

the ball must drop for 3.5 - 2.47 sec = 1.03s
find the height it falls s = 1/2 at^2
s = 0.5 x 9.8 x 1.03^2
s = 5.2m

the impact speed is 5.2m /1.03s= 5.05m/s

 

[BvU]

Hi Hal, welcome to PF :-)

Depends on whether the cliff is hit from aside or the ball lands beyond the top...

or is there an extra given bit that you missed to render ? Like the 3.5 seconds that appear from somewhere ? What's the full problem statement ?

Whatever, your last line uses an equation for average speed that cannot be applied for accelerated motion.

FInally: what about the speed in horizontal direction ?