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?
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
To solve these problems, we can use the equations of motion for projectile motion.
The equations of motion for projectile motion are:
Vertical motion equation:
h = ut + (1/2)gt^2
Horizontal motion equation:
d = ut
where:
h = height (vertical distance)
u = initial vertical velocity (m/s)
g = acceleration due to gravity (9.8 m/s^2)
t = time (s)
d = horizontal distance
a. To find the height of the platform, we need to determine the initial vertical velocity (u). Given that Phil jumps off with a horizontal velocity of 2.8 m/s and there is no information about the vertical velocity, we can assume that the initial vertical velocity is zero (u = 0 m/s).
Using the vertical motion equation:
h = ut + (1/2)gt^2
h = 0 * 2.6 + (1/2) * 9.8 * (2.6)^2
h = 0 + 12.748
h = 12.748 m
Therefore, the height of the platform is approximately 12.748 meters.
b. To find how far out from where he jumped Phil lands, we use the horizontal motion equation and the given horizontal velocity (2.8 m/s):
Using the horizontal motion equation:
d = ut
d = 2.8 * 2.6
d = 7.28 m
Therefore, Phil lands approximately 7.28 meters away from the starting point.