A rock is thrown vertically upward from the edge of a cliff. The rock reaches a

Question

A rock is thrown vertically upward from the edge of a cliff. The rock reaches a maximum height of 10.0 m above the top of the cliff before failing to the base of the cliff, landing 8.00 s after it was thrown. How high is the cliff? Between takeoff and landing, an airplane accelerates at a constant from 250 km/h to 950km/h over 30 min. Right after 30 min acceleration, the pilot realizes something wrong with the engine, instantly decelerates to stop over 20mins. Find the total distance traveled by the plane.

Explanation / Answer

(1)
h = 10.0 m

vf^2 = vi^2 + 2*a*h
0 = vi^2 - 2*9.8*10.0
vi = 14 m/s

Let the height of the cliff be h.
h = vi*t + 1/2*a*t^2
h = -14 * 8 + 1/2*9.8*8^2
h = 201.6 m

Height of the cliff, h = 201.6 m

(2)

250 km/h to 950 km/h
t = 30 min = 0.5 hr

Distance travelled during acceleration,
vf = vi + a*t
950 = 250 + a*0.5
a = 1400 km/h^2

vf^2 = vi^2 + 2*a*s1
950^2 = 250^2 + 2*1400*s1
s1 = 300 km

Distance travelled during acceleration,

vf = vi + a*t
0 = 950 - a*(20/60)
a = 2850 km/h^2

vf^2 = vi^2 + 2*a*s2
0^2 = 950^2 - 2*2850*s2
s2 = 158.3 km

Total distance travelled = s1 + s2
Total distance travelled = 300 + 158.3
Total distance travelled = 458.3 km

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