Question

A ball is thrown straight upward at an initial speed of
v0 = 80 ft/s.
(Use the formula
h = −16t2 + v0t.
If not possible, enter IMPOSSIBLE.)
(a) When does the ball initially reach a height of 64 ft?
s

(b) When does it reach a height of 128 ft?
s

(c) What is the greatest height reached by the ball?
ft

(d) When does the ball reach the highest point of its path?
s

(e) When does the ball hit the ground?

Answer 1

you have the equation h = −16t^2 + v0*t. you are also given v0, so you can plug in v0 = 80 ft/s into h to get h = -16t^2 + 80t.

for parts a, b, and e, you can plug in the respective height value into the equation and solve for t. for part a) you'd plug in h = 64, for part b) you'd plug in h = 128, and for part e) you'd plug in h = 0. in each case solve for t.

for parts c) and d) find the vertex of the equation. non-calculus method: graph, or calculate the x-coordinate of the vertex as (-b)/(2a), which will be the time at which the ball reaches the peak. plug the x-coordinate back into the equation to find the height of the ball at the peak for part d).

calculus method (if you haven't taken calculus yet don't worry about this): find the first derivative of the height equation, set it equal to 0 to get the time at which the ball reaches the peak. then plug that x-coordinate back into the original h equation to get the height for part d).