A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s.How long is the stone in the air? Assume that it lands on the ground below the cliff, and that the ground below the cliff is flat.
A)4.00 s
B)5.00 s
C)6.00 s
D)7.00 s
E)8.00 s

Question

Quizplus · Accepted Answer

We can use the kinematic equations to solve for the time the stone is in the air. Let's use the following equation:
Δy = v0*t + (1/2)*a*t^2
where Δy is the change in height (59.4 m), v0 is the initial velocity (19.5 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time. We can rearrange the equation and solve for t:
t = (-v0 ± sqrt(v0^2 - 4*(1/2)*a*Δy))/(2*(1/2)*a)
t = (-19.5 ± sqrt(19.5^2 - 4*(1/2)*(-9.81)*(59.4)))/(2*(1/2)*(-9.81))
t ≈ 5.98 s
Since we're looking for the time the stone is in the air, we'll use the positive solution for time:
t ≈ 5.98 s
So the stone is in the air for approximately 5.98 seconds, which is closest to 6.00 seconds (choice C).

A Stone Is Thrown Outward from the Top of a 59.4-M