If I am standing on top of a building that is 350 ft. high and I throw a baseball off the roof at angle of 60 degrees, determine how long the ball is in the air as well as how far it lands from the building.

Answer 1

What are being asked depends on the initial velocity.

Do you know the velocity, or you are expected to come up with an algebraic expression in terms of the initial velocity?

Is the ball thrown at 60° above the horizontal?

Answer 2

Initial velocity not given. Assuiming that we need to solve for it. Yes, the ball is thrown 60 degrees above the horizontal. Thank you.

Answer 3

Since the question asks for time in the air and distance from the building, we will use vi (initial velocity) as a parameter, and solve accordingly.
Assume:

'=feet
fps=feet per second
g=32.2 m/s²
vi=initial velocity in fps

Horizontal component of velocity, vih
=vi cos(θ)
=0.5vi fps

initial vertical position, yi
= 350'

Vertical component of initial velocity, viy
=vi sin(&theta);
=sqrt(3)/2 vi fps

Using kinematics equation to solve for t, time to reach ground
Δy=viy(t)+ (1/2)at²
-350 = sqrt(3)vi/2 - (32.2/2)t²
16.1t²-sqrt(3)vi/2-350=0
Solve for t (when vi is known).

Horizontal distance, Δx
=vix*t
=vi*t/2
where t has been solved from above.