Calculus has me stumped
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Calculus has me stumped

[From: ] [author: ] [Date: 11-09-27] [Hit: ]
Thanks for any help.-First, you need to draw the situation, labeling the distances and heights, and theangle 24 degrees.Call the initial speed V,......
Can you help me understand how to work a problem like this?

A ball hit at a 24^ angle from 3ft above ground level just clears a 9ft fence 400 ft away. About how fast was the ball moving and how long id it take to reach the fence?

I am using 32.174 ft per sec as pull of gravity.

Even a formula would be helpful. Answers will not help me if I cannot do the math. I am stumped. Thanks for any help.

-
First, you need to draw the situation, labeling the distances and heights, and the angle 24 degrees.
Call the initial speed "V", and break it up into components Vx and Vy.

Vx = V * cos( 24 deg)
Vy = V * sin( 24 deg)

The time to reach the fence is Tf = 400 / Vx.
The height of the ball at time "t" is H = 3 + Vy * t + 0.5 * g * t^2
g = -32.174 ft/ sec^2 ( <== not ft/ sec. Also notice that g is negative, because it points down.)

We are told that the ball just clears the fence. That means that if we set t = Tf, then H = 9.
9 = 3 + V*sin(24deg)*400/ (V *cos(24deg)) + 0.5(-32.174)*400^2 / (V cos(24deg))^2
You should be able to solve that for V. I get V = 133.761 ft/second, which is 91.205 mph, pretty fast.

Now that we have V, we can figure Vx and Vy, and then Tf.
(I get Tf between 3 and 5 seconds, but I'll let you figure the exact value.)
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