Question
Phil jumps off a high-diving platform with a horizontal velocity if 2.8 m/s and lands in the water 2.6 seconds later.
a. How high is the platform
b. How far out from where he jumped does he land?
a. How high is the platform
b. How far out from where he jumped does he land?
Answers
Damon
do horizontal and vertical problems separately
v = g t velocity down
h = 0 + Vo t + (1/2) g t^2 height
so
height = 4.9 (2.6)^2
height = 33.1 meters (high all right)
U = horizontal speed = constant
d = u t = 2.8 * 2.6 = 7.28 meters away
v = g t velocity down
h = 0 + Vo t + (1/2) g t^2 height
so
height = 4.9 (2.6)^2
height = 33.1 meters (high all right)
U = horizontal speed = constant
d = u t = 2.8 * 2.6 = 7.28 meters away