AP Physics z0rz w/ 1 problem

[saiyajin822]

ok I've been trying to figure out this problem for about an hour :/

A golfer drives a ball at an angle of 30 degrees with the horizontal at an initial speed of 60m/sec. How high does the ball rise?

ok I am pretty sure this is a projectile problem.
Vf=0(?)
Vh=52m/s
Vv=30m/s (i used sin and cos to get those 2 components)
g=-9.8m/s^2

I tried figuring out the time for the vertical. So, I did Vf=Vi+at. 0=30+(-9.8)t. t=3.06s. Then I tried finding the height using v=d/t. 30m/s=d/3.06. I ended up getting the wrong problem..the real answer is 45.9m.

I could realllly use some help :/

 

[saiyajin822]

nevermind, i figured it out I am just stupid ;_;

 

[M98Ranger]

Hi there,

It looks like you are on the right track with using projectile motion equations to solve this problem. However, there are a few things that may have caused you to get the wrong answer.

First, when using the equation Vf=Vi+at, make sure to use the vertical component of the initial velocity (Vv), not the horizontal component (Vh). So it would be 0=30+(-9.8)t, which gives a slightly different time of 3.06 seconds.

Second, when using the equation v=d/t, make sure to use the vertical distance traveled (since that is what we are solving for), not the horizontal distance. So it would be 30m/s=d/3.06, which gives a different answer of 91.8m.

To find the actual height, we need to use the equation d=Vit+1/2at^2, where d is the vertical distance, Vi is the initial vertical velocity, a is acceleration due to gravity, and t is time. Plugging in the values, we get d=30(3.06)+1/2(-9.8)(3.06)^2, which gives a height of 45.9m.

I hope this helps! Projectile motion problems can be tricky, so don't get discouraged. Keep practicing and you'll get the hang of it. Good luck!