ball thrown up from top of 96 foot high tower with initial velocity of 80 ft/s.
when does ball reach max height and how high is it at that time?
v = 80 - 32t
v = 0 is where the ball stops rising and starts falling
0 = 80 - 32t
t = 2.5s
h = 96 + 80t - 16t^2
= 96 + 80*2.5 - 16*6.25
= 196 ft
what is the -32 t? I think I undersaand the rest.
To determine when the ball reaches its maximum height and how high it is at that time, we can use the equations of motion for vertical motion. The equation that relates an object's vertical position, velocity, and time is:
y = y0 + v0*t - 1/2*g*t^2
Where:
- y is the vertical position at time t (height)
- y0 is the initial vertical position (96 feet)
- v0 is the initial vertical velocity (80 ft/s)
- g is the acceleration due to gravity (32.2 ft/s^2)
- t is the time
To find the time when the ball reaches its maximum height, we need to determine when its vertical velocity becomes zero. At maximum height, the ball momentarily stops before falling back down. So, we set the velocity equation equal to zero and solve for time:
v = v0 - g*t
0 = 80 - 32.2*t
Solving this equation, we find:
t = 80 / 32.2
t ≈ 2.48 seconds
Therefore, the ball reaches its maximum height after approximately 2.48 seconds.
To determine the height at that time, we substitute this value of time back into the equation that relates the vertical position and time:
y = y0 + v0*t - 1/2*g*t^2
y = 96 + 80*2.48 - 1/2*32.2*(2.48)^2
y ≈ 146.6 feet
So, at the maximum height, the ball reaches approximately 146.6 feet above its initial position.