$1.$ A ball is hit with an initial horizontal velocity of 33 ft/sec and initial vertical velocity of 51 ft/ sec. How long does it take for the ball to go from its max height to the ground?
My attempt:
I assume I only need to worry about vertical velocity to solve this and gravity is -32 ft/sec. Thus I determined:
$\int -32 dt = v_y = -32t + c$
Given the initial vertical velocity of 33, then $v_y = -32t + 33$. Then to get vertical distance we integrate again which gives us $-16t^2 + 33t = d_y$ (y distance). Plug $0$ in for $d_y$ and you get $t=1.44$. I'm not sure if this is the correct time though. Is this just the total time the ball is in flight?
$2.$ A ball takes 3 seconds to travel from its max height to the ground. The initial horizontal velocity of the ball is 27 m/s.
a. What is the angle at which the ball takes off?
b. What is the initial vertical velocity of the ball?
If I could find the initial vertical velocity, I could do $arctan(v_{yo}/v_{xo})$, but not sure how to find it. Thanks for any help 🙂
Best Answer
For $1$:
yes, that's the total time it's in flight. (Also, you should have a definite integral:
$$v_f - v_i = \int_{t_i}^{t_f} -32~dt = -32t \implies v_f = -32t + v_i = -32t + 33$$
For $2$:
You're right about finding the initial vertical velocity first. You know that $-32 t + v_{y,i} = 0$ when $t = 3$ (because at its apex, there is no vertical motion), so can you find $v_{y,i}$?
I suppose you could also model it as a ball fired from a certain (unknown) height, and than after $3$ seconds, you declare that to be the ground. Then find the angle it's moving at, but that's not as natural a solution. (Especially because either way, you compute $v_{y,i}$ first.)