A girl on a trampoline jumps up with a velocity of 5.0 m/s . Using the quadricti

Question

A girl on a trampoline jumps up with a velocity of 5.0 m/s. Using the quadrictic equation calculate how high the girl jumps, and the total length of time she will be in the air.

- 9.8m/s2 = a = g; y0 = 0, v=0, v0 = 5.0m/2
equation1: v2 = V02 + 2g (Y - Y0);    equation 2: y= y0 + v0 + 1/2gt2

First find Y with equation 1, Y will tell you how high the person goes
Then using the data from equation 1 solve equation 2 to find time.

BUT..... the time in equation two is only for the time in air on way up. I need help finding total time in air. AND I need to solve the equation using the quadratic equation.

In answering this question please show all steps taken to solve, and not just the answer. As the point of my question is to understand how to do a problem like this, not simply to see an answer.

Explanation / Answer

For equation 1, with the given values, we have
0^2=5^2 + 2 (-9.8) (Y-0)
Y=1.3m (how high the girl jumps)

For equation 2 (your equation 2 is wrong), y=y0 + v0t + 1/2 gt^2
With the given values, we have
1.28=0 + 5t + 1/2 (-9.8) t^2
4.9t^2 -5t + 1.28=0 with a=4.9, b= -5, c=1.28
t= -b±(b^2-4ac)/2a = 5±[25-4(4.9)(1.28)]/2(4.9)=0.51s

So this is time for way up. Time for way up=time for way down, so 0.51x2=1.0s

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