# Projectile Launched Question

## Homework Statement

A 500kg object is launched from the top of a 12m building with intial speed of 5.2m/s and angle of 38degrees. How far away does object land?

y=vit + 1/2at^2
d=vt

## The Attempt at a Solution

first i separate the initial velocity into its x and y coordinates. Using cos and sin. Next I input this found value of the y direction velocity into the y=vit + 1/2at^2. I rearrange this formula and solve it as a quadratic for time, and I use the positive time value which I got as 1.92s. Next I multiply this time value by the velocity in the x direction I got from doing the cos of the initial velocity.

I end up getting the object lands 7.9 m away from the building. Can anyone check and see if this is right? Also the mass was negligble right?

## Answers and Replies

Your method is correct.

If you want your numbers to be checked, post the intermediate values you got (init. velocity components, time of flight).

Your method is correct.

If you want your numbers to be checked, post the intermediate values you got (init. velocity components, time of flight).

Well the initial velocity was 5.2m/s. The y component of this is 3.2m/s, and the x component is 4.097 m/s.

When I solve for time from the quadratic formula, I get -1.272s and 1.923s. I use the positive value cause you cant have negative time.

Now I just multiply the time by the x component velocity and get 7.9m.

Everything looks good.

1 person
Everything looks good.

Thanks.