- Chen
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Since the cliff is 80-meter-high, we can easily figure out the fallen time of the shell is 4 seconds. (we can work out that variable by the formula, (1/2)*gt^2=h. In that formula, "g" stands for acceleration of gravity, "t" is time, and "h" means the high above from the ground).
In the horizontal direction, the shell moves at a uniform speed in a straight line. So the distance between the fallen shell and the base of the cliff (1330 m) indicates how fast the shell was at the very beginning when it was fired.
Vx=s/t=1330/4=332.5 (m/s); Vx=horizontal speed; s=distance=1330 m; t=fallen time=4 s.
The shell fell down the whole process under the gravitation. Therefor the acceleration of the shell is always 9.8 m/s^2 wherever it is. (usually we treat that as 10 m/s^2)
The shell moves in 2 directions, the horizontal one and the vertical one. The horizontal speed is 332.5 m/s, which we got just now, and the vertical speed could be worked out by that formula, Vy=gt=10*4=40 m/s; In here, Vy=vertical speed (in the moment just before it strikes the ground); g=acceleration of gravity=10 m/s^2; t=fallen time=4 s;
So the speed final is V=sqrt(Vx^2+Vy^2)=√ˉ(332.5^2+40^2)=334.897372339647 m/s.
We can answer the 2 questions now.
The magnitude of the acceleration of the shell just before it strikes the ground is 9.8 m/s^2, but usually we treat that as 10 m/s^2 for convenience.
The speed of the shell as it hits the ground is 334.897372339647 m/s. It composes by the horizontal speed and the vertical speed.
- gitcloud
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what is the magnitude of the acceleration of the shell just before it strikes the ground?
9.8m/s^2
Vy^2 = 2gh
Vy = (2*9.8*80)^(1/2)
t = Vy/g = (2*80/9.8)^(1/2)
Vx = 1330/t = 1330*(9.8/160)^(1/2)
V
= (Vx^2+Vy^2)^(1/2)
= (1330^2*9.8/160+2*9.8*80)^(1/2)
= 332m/s
what is the speed of the shell as it hits the ground?
332m/s