
Arturo O. answered 03/07/18
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Solve this as a projectile problem with vertical motion at constant acceleration.
h(t) = -gt2/2 + v0t + h0
g = 9.8 m/s2
v0 = 12 m/s
h0 = 0
h(t) = -4.9t2 + 12t
v(t) = -gt + v0 = -9.8t + 12
Maximum height occurs when v(t) is instantaneously zero.
0 = -9.8t + 12
t = 12/9.8 s ≅ 1.224 s
Evaluate h(t) at t = 1.224 s to get hmax. You can finish from here.

Arturo O.
You are welcome, Christopher.
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03/07/18
Christopher N.
03/07/18