What is the initial velocity of a home run ball?

[Jameson]

I need help!

Some smart physics person, help me out here.

A ball player hits a home run, and the baseball just clears a wall 18.5 m high located 140.0 m from home plate. The ball is hit at an angle of 32° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.

a. What is the initial speed of the ball? Solve for time in the x-direction and substitute it in the formula for the vertical position.


b. How much time does it take for the ball to reach the wall? Solve using only the vertical component of the velocity or the horizontal component.


c. Find the velocity components and the speed of the ball when it reaches the wall.
Vy,f
Vx,f
Vf

 

[futb0l]

the first one is pretty simple - the question already tells u what to do.

use this equation and get t:

$$x = (v_i \cos (\alpha))t$$

once u get t, u can sub it in

$$y = y_i + (v_i \sin (\alpha))t - \frac{1}{2}gt^2$$

and u will get the initial speed.

Well - u already got the time - but they want u to get it off only the vertical component - so all u do is put y = height of the wall and since u have all teh other variables - u can now solve for t.

Now for the velocities ... use the equation and sub in t:

$$v_y = v_i \sin (\alpha) - gt$$

$$v_x = v_i \cos (\alpha)$$

i don't know about v_f - but i think u shuld use the equations from above and find y(x) and get the derivative at time t, actually i think u can just vectorially do Vy + Vx

 

[Jameson]

I don't know how to solve for t... I'm sorry this seems dumb. But I need more help

 

[thermodynamicaldude]

...x = v cos (theta) t

divide both sides by vcos(theta)...and you get

t = x/(v cos (theta)

Now plug this into the equation for y...

When you do this, you should be able to solve for initial velocity.