Projectile prob: ball kicked from ground at angle above...

[jwpaine]

A ball kicked from the ground at an angle of b above the horizontal with an initial speed of \(\displaystyle V_0\) feet per second reaches a height of \(\displaystyle \L h(t) = V_0 \cdot t \cdot sin(b) - \frac{1}{2} \cdot g \cdot t^{2}\)

gravity = g = 32 feet per second
angle = b = 30 degrees

The ball reaches a max height of 16 feet.

What was the initial velocity?

\(\displaystyle \L h(t) = V_0 \cdot t \cdot sin(\frac{\pi}{6}) - \frac{1}{2} \cdot 32 \cdot t^{2}\)

\(\displaystyle \L h(t) = \frac{1}{2}\cdot V_0 \cdot t - 16t^2\)

Now I know that when h(t) = 16 that the derivitive of the function has slope = 0, because it's at the vertex... but I'm just not sure how to find the initial velocity and time with that one equation.

I should know this from high school physics

 

[TchrWill]

Re: Projectile prob: ball kicked from ground at angle above.

A ball kicked from the ground at an angle of b above the horizontal with an initial speed of \(\displaystyle V_0\) feet per second reaches a height of \(\displaystyle \L h(t) = V_0 \cdot t \cdot sin(b) - \frac{1}{2} \cdot g \cdot t^{2}\)

gravity = g = 32 feet per second
angle = b = 30 degrees

The ball reaches a max height of 16 feet.

What was the initial velocity?

\(\displaystyle \L h(t) = V_0 \cdot t \cdot sin(\frac{\pi}{6}) - \frac{1}{2} \cdot 32 \cdot t^{2}\)

\(\displaystyle \L h(t) = \frac{1}{2}\cdot V_0 \cdot t - 16t^2\)

Now I know that when h(t) = 16 that the derivitive of the function has slope = 0, because it's at the vertex... but I'm just not sure how to find the initial velocity and time with that one equation.

The height reached comes from h = Vo^2(sin^2(µ))/2g or h = (gt^2)/2.

Therefore, 16 = Vo^2(sin^2(30))/2(32) = 4096 making Vo = 64 fps.

Since 16 = 32(t^2)/2, t^2 = 2(16)/32 = 1 making t = 1 sec.

Checking with h = Vv(t) - gt^2/2 = 64(sin30)(1) - 32(1)^2/2 = 32 - 16 = 16 ft.

 

[jwpaine]

how did you get h = Vo^2(sin^2(µ))/2g

why is initial velocity and sin squared? and why over 2g? I'm guessing you went backwards from a derivative.

 

[TchrWill]

how did you get h = Vo^2(sin^2(µ))/2g

why is initial velocity and sin squared? and why over 2g? I'm guessing you went backwards from a derivative.

The 3 basic equations of projectile motion are
1--Vf = Vo -gt
2--h = Vot - gt^2/2
3--Vf^2 = Vo^2 - 2gh

Consider the initial horizontal and vertical components of the initial velocity Vo where Vh = Vo(cos(µ)) and Vv = Vo(sin(µ)).
The horizontal distance traveled, or the range, is given by d = Vo(cosµ)t1.
The time to maximum height derives from (1) where Vf = 0 and Vfsinµ = gt1 whence t = Vfsinµ/g.
During this period of t1, the projectile rises to a height given by (2) as h = (Vfsinµ)t1 - gt1^2/2 which can be written as h = gt1^2 - gt1^2/2 = gt1^2/2.

The time, t2, for the projectile to return to the ground is derived from -h = -gt2^2/2. It follows that the distance up equals the distance down and t1 = t2, the total time of flight being t = 2t1. Combining, d = Votcosµ = 2(vo)t1(cosµ) = 2vo(Vo(sinµ))cosµ/g) or d = (Vf^2)(sin2µ)/g.

Eliminating t1 from Vo(sinµ) = gt1 and h = g(t1^2)/2, the maximum height reached is given by h = Vo^2(sin^2µ)/2g