Physics question.Please help!
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Physics question.Please help!

[From: ] [author: ] [Date: 12-05-19] [Hit: ]
9*t^2 => t = sqrt[2*72.97/9.8] = 3.1.27+3.86 = 4.......
A stone is thrown vertically upward with a speed of 12.5 m/s from the edge of a cliff 65.0 m high (Fig. 2-34).

(a) How much later does it reach the bottom of the cliff?

(b) What is its speed just before hitting?

(c) What total distance did it travel?

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The vertical motion Vy(t) = Vyi -g*t = 0 at the moment of peak altitude => Vyi/g = t
t = 12.5/9.81 = 1.27s

Use Vf^2 = Vi^2 + 2a*d to find the peak altitude when Vf = 0, a = -g and d = peak altitude
d = Vi^2/2g = 12.5^2/19.6 = 7.97m

To find the time to reach ground use d = do + Vi*t + 1/2 *a*t^2 where in this case do = 0, Vi = 0,a = g and
d = 65+7.97 = 72.97m

72.97 = 4.9*t^2 => t = sqrt[2*72.97/9.8] = 3.86s

Total time aloft is time to peak height plus time to ground

1.27+3.86 = 4.92s <----------------------------- (a)

Use V(t) = Vi + a*t = 0 + g*3.86 = 37.81m/s <----------------------------- (b)

Total distance traveled is up to peak altitude plus distance back to launch height plus 65m to ground

7.97+7.97+65 = 80.94m <----------------------------- (c)

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A stone is thrown vertically upward with a speed of 12.5 m/s from the edge of a cliff 65.0 m high (Fig. 2-34).

(a) How much later does it reach the bottom of the cliff?
Distance = vi * t + ½ * g * t^2
65 = 12.5 * t + ½ * 9.8 * t^2
Solve this quadratic equation for time.
4.9 * t^2 + 12.5 * t – 65 = 0
t = [-12.5 ± (12.5^2 – 4 * 4.9 * -65)^0.5] ÷ 9.8
(12.5^2 – 4 * 4.9 * -65)^0.5 = 37.82
t = [-12.5 + 37.82] ÷ 9.8 = 2.58
t = [-12.5 – 37.82] ÷ 9.8 = -5.13 (neglect negative time)

(b) What is its speed just before hitting?
As the stone fell for 2.58 seconds, its velocity increased 9.8 m/s each second.
Final velocity = Initial velocity + 9.8 * t = 12.5 + 9.8 * 2.58 = 37.784 m/s

(c) What total distance did it travel?
The stone fell from the edge of a cliff 65.0 meters high to the ground, which is 65 meters below the edge of a cliff. So I believe the stone fell 65 meters!

Distance = 12.5 * 2.58 + ½ * 9.8 * 2.58^2 ≈ 64.9 m
The error is to rounding!

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vo= 12.5m/s
a=gravitityconstant: g=9.8 m/s^2
height of the cliff: 65.0 m

use the formula: s(t)=-g t^2 /2 + vo t + 65.0
=-4.9 t^2 + 12.5 t +65.0

s(t)=0

0=-4.9 t^2 + 12.5 t +65.0
solve for t=5,13456

b) derivate s(t): v(t)= ds(t)/dt = -9.8 t + 12.5
and
v(5.13456)= -37.81
1
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