How High Does a Golf Ball Rise When Hit at 30 Degrees?

AI Thread Summary
A golfer hits a ball at a 30-degree angle with an initial speed of 60 m/s, prompting a discussion on how high the ball rises. The vertical component of the initial velocity is calculated as 30 m/s, and the acceleration due to gravity is -9.8 m/s². The correct method involves using the equation d = Vit + 1/2at² to find the height, leading to a maximum height of 45.9 m. The initial confusion stemmed from misapplying the equations related to vertical and horizontal components. Understanding projectile motion equations is essential for solving such problems accurately.
saiyajin822
Messages
17
Reaction score
0
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 :/
 
Physics news on Phys.org
nevermind, i figured it out I am just stupid ;_;
 


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!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top