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## Homework Statement

The starting hight is .51 metres above the ground, and the range is 4.651 metres. the angle of launch is 22 degrees above the horazontal. i need to find the initial velocity, and this was all i have to work with.

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- Thread starter xwater
- Start date

- #1

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The starting hight is .51 metres above the ground, and the range is 4.651 metres. the angle of launch is 22 degrees above the horazontal. i need to find the initial velocity, and this was all i have to work with.

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tiny-tim

Science Advisor

Homework Helper

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Hi xwater ! Welcome to PF!

Start by calling the speed v, and writing (separate) equations for the x and y components of the motion.

What do you get?

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- #5

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In the x direction you have vcos22. In the y direction you have vsin22.

Now start using kinematics.

The x direction is simple. For a total flight time t,

t = 4.651/vcos22

The y direction is slightly more complicated. The projectile's path goes up, then back down to where it started and then continues for another 0.51 meters. This means its final displacement is 0.51 meters below its initial position. So,

-0.51 = vsin22t - 0.5gt^2

But we know t from before.

-0.51 = vsin22*4.651/vcos22 - 0.5g(4.651/vcos22)^2

-0.51 = tan22*4.651 - 0.5g(4.651/vcos22)^2

-0.51 - tan22*4.651 = - 0.5g(4.651/vcos22)^2

[-0.51 - tan22*4.651]/(-0.5g) = (4.651/vcos22)^2

And I think you can see where it goes from there.

- #6

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The explaination by Atticus is nice and correct! :) I second it! :)

- #7

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Thanks a ton guys! Atticus, you blew my mind..lol. Now i just gotta go do it myself :)

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Shouldn't "-0.51 = vsin22t - 0.5gt^2" be -0.51 = vsin22t + 0.5gt^2?

The formula is d = Vit + 1/2 at^2 is it not?

- #9

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Shouldn't "-0.51 = vsin22t - 0.5gt^2" be -0.51 = vsin22t + 0.5gt^2?

The formula is d = Vit + 1/2 at^2 is it not?

In a problem where the net acceleration is in a direction opposite to the positive direction you chose, the "a" in your formula becomes negative.

In other words, when you move in the positive direction (in this case upwards) you slow down, when you move in the negative direction (downwards) you speed up.

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