At what angle should we launch a projectile in order to maximize the vertical height it obtains?

Question

Use the kinmatic equations to calculate the vertical hieght obtained by a golf ball in yards that is hit at this launch angle with an initial speed of 90 miles per hour.

Answer 1

vertical height? without using kinematic equations i can tell you that since vertical and horizontal motion are independent vectors and since only the vertical vector is affected by gravity you would logically want to reduce velocity taken up in the horizontal vector by shooting it at any angle other than 90 degrees. so you'd launch the projectile at 90 degrees to maximize the vertical height... :P

don't listen to the person below either... he's daft... http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
will demonstrate that 90 degrees is the optimum angle to maximize the vertical height... i'll leave the rest of your homework to you....

Answer 2

Assuming no air resistance using kinematic equations 90miles/hour = 144.84km/hour
144.84km/hour = 40.2333m/s
optimum angle = 45 degrees
sin 90 = 1
the vertical componet of the speed = 40.2333m/s * sin90
= 40.2333m/s upwards
Vf = Vi + A (T)
0 = 40.2333m/s - 9.81 t
t = 4.10sec
average speed of 20.12m/s (40.233/2)
20.12m/s * 4.1s = 82.49m = maximum height
82.49m = 0.0512 miles = 270.6 feet

45 degrees is the best angle if you wish to send the projectile the furthest vertically and horizontally

Answer 3

Straight up. Don't waste energy on getting to the green. Bad golf, but then I'm not sure a lot of physicists play golf.