[SOLVED] Projectile Motion

Apr 2010
29
0
Hi

Gee Ming the golfer hits a ball from level ground with an initial speed of 50m/s and an initial angle of elevation of 45 degrees. The ball rebounds off an advertising hoarding 75 m away. Take g = 10m/s^2

a) Show that the ball hits the hoarding after 3/2 * sqrt(2) s at a pt 52.5 m high.

b) Show that the speed v of the ball when it strikes the hoarding is 5 sqrt(58) m/s at an angle of elevation alpha to the horizontal, where alpha = arctan (2/5).

c) Assuming that the ball rebounds off the hoarding at an angle of elevation alpha witha speed of 20% of v, find how far from Gee ming the ball lands.

How exactly do i do c???

By the way, the answer's meant to be 50 m
 
Last edited by a moderator:
May 2010
20
8
Hi

Gee Ming the golfer hits a ball from level ground with an initial speed of 50m/s and an initial angle of elevation of 45 degrees. The ball rebounds off an advertising hoarding 75 m away. Take g = 10m/s^2

a) Show that the ball hits the hoarding after 3/2 * sqrt(2) s at a pt 52.5 m high.

b) Show that the speed v of the ball when it strikes the hoarding is 5 sqrt(58) m/s at an angle of elevation alpha to the horizontal, where alpha = arctan (2/5).

c) Assuming that the ball rebounds off the hoarding at an angle of elevation alpha witha speed of 20% of v, find how far from Gee ming the ball lands.

How exactly do i do c???

By the way, the answer's meant to be 50 m

For part a, \(\displaystyle v_o=50m/s\) at \(\displaystyle 45^o\) therefore,

\(\displaystyle v_{o_x}=25m/s\) and \(\displaystyle v_{o_y}=25m/s\)

\(\displaystyle v_{o_x}\) is constant so the time it takes the ball to travel to the sign should be given by the equation,

\(\displaystyle time=\frac{distance}{velocity}=\frac{75m}{25\frac{m}{s}}=3s\)

which is pretty close to \(\displaystyle \frac{3\sqrt{2}}{2}\)

am I missing something here?
 
Apr 2010
29
0
Sorry for the inconvenience but I've done both a + b. I just dont know how to do c. (Crying)