Physics: projectile motion: maximum height

Continuing about projectiles, the tutor finds the maximum height of the golf ball.

My post from Oct 16 sets the premise: a golf ball is struck at 55m/s at 27° elevation. The velocity is resolved thus:

vx=49m/s

vy=25m/s

To find the maximum height, we focus on vy. Specifically, when vy=0, the ball is at max height.

We can use the formula

vf^2=vi^2+2ad

in which vf is the final velocity; vi , the initial; a, the acceleration due to gravity; d, the displacement (in this case, the vertical displacement: the height).

0^2=25^2+2(-9.8)d

0=625 -19.6d

19.6d=625

d=625/19.6=32m

The golf ball reaches max height of 32m.

HTH:)

Source:

Giancoli, Douglas C. Physics. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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