A golf ball is hit with an initial angle of 36.9o with respect to the horizontal and an initial velocity of 83

Question

A golf ball is hit with an initial angle of 36.9o with respect to the horizontal and an initial velocity of 83.9 mph. It lands a distance of 89.5 m away from where it was hit. By how much did the wind resistance reduce the range of the golf ball?

Answer 1

We can derive the equation...

R=(2vi^2sin2x)/g

R=range
x=angle
g=gravity

...using motion in 2 dimensions. If you want to see how I did this, then email me.

Now, all that we have to do is plug in our angle, initial velocity, and gravity to find our answer.

I hope that this helps.

Answer 2

First, how long will the ball proceed to be in the air? Its vertical area of action would be approximately 33 m/s, so use the equation 0 = 33 m/s + (-9.8 m/s²) t. The ball will attain its optimal element at t (3.4 s) and could hit the floor at 2t (6.8 s). From there, you will locate horizontal displacement with the equation ?x = (60 m/s * sin 33°) (6.8 s). Your answer would be 340 m taking substantial figures under consideration. extremely a force!

Answer 3

just calculate theoretical distance and subtract 89.5. to get theoretical use cosine to get velocity in x direction. have to convert mph to m/s. turns out to be 37.51m/s. v turns out to be 30. then you have to find time it takes in the y direction to get up and down. find v in y direction(turns out to be 22.52) then use vf=vi+at. 0=22.52+(-9.8)t. turns out to be 2.3 seconds. times by two for up and down to get 4.6 seconds. then use s=.5at^2+vit with the horizontal velocity we got earlier (a being 0 in x direction) to get s to be 137.88m. subtract the 89.5 to get 48.38m lost to wind resistance