Physics help





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Discus: SAT/ACT Tests and Test Preparation: October 2003 Archive: Physics help
By Euphoria (Euphoria) on Thursday, October 02, 2003 - 06:13 pm: Edit

A stone is thrown vertically upward with a speed of 10 m/s from the edge of the cliff 65 m high. It reaches the ground sometime later

a) what is the acceleration and velocity at the highest point. (I got acceleration=9.8 m/s^2 and velocity=o)

b)What is the displacement for the freefall portion of the trip? (I got 0 m)

c) what is the final free fall velocity?

d) how long is the stone in freefall?

2)A car starts from rest and accelerates uniformily to a speed of 30 m/s in 10s. The driver maintains this speed for another 10s.

What is the displacement for the entire 20s trip?


Thanx a lot

By Clickspring (Clickspring) on Thursday, October 02, 2003 - 08:20 pm: Edit

The displacement is 65m because it goes from the top of the cliff 65m above the ground, which is at 0.

By Perry2006 (Perry2006) on Thursday, October 02, 2003 - 10:18 pm: Edit

a) what is the acceleration and velocity at the highest point. (I got acceleration=9.8 m/s^2 and velocity=o)
Acceleration = -9.8 m/s^2 and velocity = 0

b)What is the displacement for the freefall portion of the trip? (I got 0 m)
-65 m

c) what is the final free fall velocity?
That is the same as saying what is the final velocity (Vf)
Vf^2 = Vi^2 + 2aX
Vf^2 = 10^2 + 2(-9.81)(-65)
Solve for Vf, you have -37.085 = -37 m/s

d) how long is the stone in freefall?
X = Vi(t) + .5(-9.81)(t)^2
-65 = 10(t) + .5(-9.81)(t)^2
Solve for t, you have 4.7997 s = 4.8 s

2)A car starts from rest and accelerates uniformily to a speed of 30 m/s in 10s. The driver maintains this speed for another 10s.

What is the displacement for the entire 20s trip?
First of all, solve for displacement of the first one.
X = .5 (Vi + Vf) (t)
X = .5 (30)(10)
X = 150 m

Second trip, the displacement will be:
X = V * t
X = 30 * 10
X = 300

Thus the total displacement is 300 + 150 = 450 m.


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