Finding the max height of rock launched by catapult

[espinosav1]

The question is: a catapult lauches a rock at 30 degree angle with respect to the horizontal. Find the maximum height attained of the speed of the rock at its hightest point is 30 m/s.

I have tried a bunch of different formulas. I'm not sure if I should use the y or x component of the equation.

I've been using v^2= (vcos(angle))^2 + 2gh

I know I'm supposed to use some form of this equation but I'm stuck.
Help please!
Thanks

 

[TchrWill]

Re: Finding the max height

The question is: a catapult lauches a rock at 30 degree angle with respect to the horizontal. Find the maximum height attained if the speed of the rock at its hightest point is 30 m/s.

The two projectile motion expressions that define the trajectory of the projectile are

h = V^2(sin^2µ)/2g

d = V^2(sin2µ)/g

where h = the maximum height of the projectile, d == the horizontal distance traveled by the projectile, V = the initial velocity, g = 9.8m//s^2 = the acceleration due to gravit and µ = the angle of the projection velocity to the horizontal.

These derive from breaking the initial projection velocity into its vertical and horizontal components. Vv = Vvertical = Vsinµ and Vh = Vhorizontal = Vcosµ.

The time to maximum height derives from Vf = Vv - gt where Vf = the final vertical velocity or 0 = Vv - 9.8t or t = Vv/9.8. The time from the maximum height back to the ground is the same time as it took to climb to its maximum height making the total time of flight T = 2Vvsinµ.

The height reached derives from h = gt1^2/2.

The horizontal velocity component remains constant over the rise and fall time which is given as 30m/s. Therefore, V = 30/cos30 = 34.64m/s.

Therefore, Vv = 34.64sin30 = 17.32m/s from which t1 = 17.32/9.8 = 1.767 sec. making T = 3.534 sec.

Therefore, h(max) = 9.8(1.767^2)/2 = 15.30m.

Using the combined equation, h = (34.64^2)sin^2(30)/2(9.8) = 15.3m.




Thanks

 

[espinosav1]

the problem

Hi
I just want to make sure i did it right.
If the angle was changed to 48 and the velocity to 31.1
would the max height by 60.86.
Thanks
Vanessa

 

[TchrWill]

Re: the problem

Hi
I just want to make sure i did it right.
If the angle was changed to 48 and the velocity to 31.1 would the max height by 60.86.

The horizontal velocity component remains constant over the rise and fall time which is given as 31.1m/s. Therefore, V = 31.1/cos48 = 46.478m/s.


Vv = 46.478sin48 = 34.54m/s from which t1 = 34.54/9.8 = 3.524 sec. making T = 7.048 sec.

Therefore, h(max) = 9.8(3.524^2)/2 = 60.86m.

Right on!