6. A stone is dropped from the upper observation deck of a tower, 50 m above the

Question

6. A stone is dropped from the upper observation deck of a tower, 50 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 7 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

Explanation / Answer

h = 50 m

a = g = 9.81 m/sec^2

Vi = 0

distance traveled in time t, will be:

H = Vi*t + 0.5*g*t^2

H = 0*t + 0.5*9.81*t^2

distance from ground will be

h(t) = 50 - H

h(t) = 50 - 4.905*t^2

B.

50 = 0*t + 0.5*9.81*t^2

t = sqrt (2*50/9.81) = 3.19 sec

t = 3.19 sec

C.

Vf^2 = Vi^2 + 2*a*H

Vf = sqrt (0^2 + 2*9.81*50) = 31.32 m/sec

Vf = 31.32 m/sec

D.

Vi = 7 m/sec

H = Vi*t + 0.5*g*t^2

50 = 7*t + 0.5*9.81*t^2

4.905*t^2 + 7*t - 50 = 0

Solving above quadratic equation:

t = 2.56 sec

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