A golf ball is it with the initial velocity of 50.0 m/s at an angle of 55 degrees above the horizontal. (a) How high will the ball go? (b) What is the total time the ball is in the air? (c) How far will the ball travel horizontally before it hits the ground?

Answer 1

STEP 1:
Determine horizontal(x) and vertical (y) components of the initial velocity of golf ball.
Vy=Vsin55 = 50x.8191= 40.9 m/s, Vx=Vcos55 = 50x.5735= 28.6 m/s

STEP 2:
At maximum height, Vy=0.
Determine time (t) it takes to reach max. height.
Vy(final)=Vy(initial)+(-g)x(t)
Solve for (t):
0=40.9 x (-9.8) x t
0=40.9 x (-9.8 t)
9.8t=40.9, t=40.9/9.8, t= 4.17 sec
Since this half the total time in air, multiply by 2,
= 8.3 SEC (TOTAL T IN AIR)
STEP 3:
DETERMINE HIGHEST POINT REACHED BY BALL
h=((V**2)(Sin**2 55))/ 2g
h=((50**2) (.8191**2)) / 2 (9.8)
h=((2500)(.6710))/19.6
h= 85m Height reached

STEP 4:
DETERMINE 'RANGE' (total horizontal distance traveled)
R=Vx (t)
R=28.6 x 8.3
R=237.38 meters

Answer 2

thank you very much for your help M Bil