Physics Question?

Question

A golf ball is struck with a five iron on level ground. It lands 105.0 m away 5.00 s later. What was the magnitude and direction of the initial velocity? (Neglect air resistance.)

____________ (m/s)
____________ * (Above the Horizontal)

Answer 1

2.50 seconds after the hit, y-velocity has gone to 0

since y-velocity decreases at 9.8 m/s^2, so 9.8 * 2.5 is the original y-velocity, and 105.0 / 5.00 is the x-velocity

velocity is square root of (x^2 + y^2)

Answer 2

You know that the horizontal component of voh is 105/5 or 21 m/sec. The next step is to figure out the vertical component.

Parabolic trajectory curves are symmetrical, so the ball took 2.5 seconds to fall to the ground. You can backtrack to find the height, and since
x =vov*t - 4.9 t2 on the way up , where t=2.5 sec and x=height from x= 4.9*2.5^2, you can solve for vov, which is the y component of initial velocity. Then the initial velocity is sqrt(vov^2+voh^2) and the angle is tan-1 (vov/voh)

Answer 3

I always like to use the symmery of projectile problems to cut through the calculation. So if the ball flies for 5 sec , it spends 2.5 secons going up and 2.5 sec going down. At the high point, the ball has 0 m/s speed in the y (vertical) direction. So using

v = v0y +at where v0y is the initial speed in y direction and a = -g = -9.8 m/s^2 (miuns sign because I picked "down" to be in negative y direction). Setting v = 0, t =2.5 and

v0y = v0*sin(q) --> q = angle initial velocity make with horizontal (x) direction.

0 = v0*sin(q) - 9.8 *(2.5)^2 --> v0*sin(q) = 61.25

v0 = 61.25/sin(q)

Now we need another equation since v0 and sin(q) are unknown. That would be the distance traveled in the x (horizontal) direction:

x = v0x*t = v0*cos(q) *2.5 = 105

105 = 61.25/sin(q) *cos(q) *2.5 -->cot(q) =0.6857

or tan(q) = 1.45833 ---> q = 55.56 deg

Since v0=61.25/sin(q) = 74.27 m/s