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RJLiberator

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## Homework Statement

A soccer ball is kicked from the ground at an angle of 30◦ with speed 30m/s. Determine the

duration of the flight, range and maximal height of the trajectory.

## Homework Equations

## The Attempt at a Solution

1. Break down the velocity vector into x and y components:

30cos(30) = 15sqrt(3) m/s

30sin(30) = 15m/s

r(t) = (0,0)+t(15sqrt(3), 15)+(t^2)/2(0,-9.80)

This equation represents the position function.

If I take the derivative to get the velocity function, it becomes

r'(t) = (15sqrt(3), 15-9.8t)

When velocity in the y direction is 0, then the ball is at max height

so 15-9.8t=0

t=15/9.8 for max height.

When position of y = 0 the ball reaches its distance and final duration so:

15t-4.9t^2=0

t=0 or t= 15/4.9

clearly it is not t=0.

since we have time of duration:

t*velocity in x = range

15/4.9*15sqrt(3) = 79.5329 m

max height is found by dividing the time of max range by 2 so thew time of max height = 15/9.8 and then using that as t in the r(t) position function and finding the y-component.

The y-component comes out to be 11.48m. Seems reasonable.

15*15/98-9.8/2*(15/9.8)^2=11.48m

Am I doing this right? My problem is that our calculus class requires no calculators. To me, it seems like this problem is begging the use of calculators unless I just don't simplify things (which would be fine to ). It just seems like I am missing something that'd make the calculations easier.

Thanks.

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