Projectiles question?

Question

This problem has confused me for awhile now.

A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high was the tallest spot in the ball's path?

Answer 1

First answer was so long I went for the math-only form.
vy = vsin(25deg)
vx = vcos(25deg)
t = x/vx, where x = 42 m
hmax = vy^2/2g
The answers are the same as 1st answer, except without the multiple steps and rounding and still using g = 9.8, t = 2.015 s and hmax = 4.821 m.

Answer 2

It's been many years since I did these kind of problems, but I think I've refreshed my memory enough to help you out now...

d - distance
v - velocity or speed
t - time
a - acceleration (gravity = 9.8m/s^2 down)

dx - horizontal distance
vx - horizontal speed (a constant)
dy - vertical distance
vy - INITIAL vertical speed (it changes due to gravity)
vy2 - vertical speed at time t
We know that distance travelled is equivalent to speed x time:
d = vt
Therefore:
dx =(vx) t
Rearranging to solve for 't':
t = dx / vx
t = 42.0m / vx

To find vx you have to use your initial speed and it's angle. See: http://www.makingthemodernworld.org.uk/learning_modules/maths/04.TU.02/illustrations/04.IL.04.gif

So...
vx = 23.0 m/s cos (25)

Plug this into the above equation:
t = 42.0m / ((23.0 m/s) (cos (25)))
t = 2.01s

Therefore the ball was in the air for 2.01 seconds.

To figure out how high the ball travelled you need to figure out when (t) the ball reached the top of it's arc. This is the point at which it stops travelling up and will start coming back down...that means that at the very top of the arc the vertical velocity is zero (vy2=0).

vy2 = vy - gt
Rearrange to solve for t:
t = (vy2 -vy) / -g
t = (0 - ((23.0m/s)sin(25)) / (-9.8m/s^2)
t = 0.99 s

So 0.99s into the ball's travel, it reaches it's maximum height. To find what this height is, plug your known values into the following equation:
dy = (vy)t + 1/2 a (t^2)
dy = ((23.0m/s)sin(25)) + 1/2(-9.8m/s^2)(0.99^2)
dy = 4.9 m

Hopefully someone else can verify my answers for you since I am admittedly rusty with this stuff :)