Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)
A ball is thrown vertically upward from a height of 4 feet with an initial velocity of 79 feet per second. How high will the ball go? (Round your answer to two decimal places.)
a(t) = -32
v(t) = 79 - 32t
h(t) = 4 + 79t - 16t^2
Now just find the vertex of that parabola, at t = 79/32
To find the maximum height the ball will reach, we need to determine when its velocity becomes zero.
Given:
Initial height (s₀) = 4 ft
Initial velocity (v₀) = 79 ft/sec
Acceleration due to gravity (a(t)) = -32 ft/sec²
We can use the kinematic equation for displacement in terms of initial velocity, final velocity, and acceleration:
v² = v₀² + 2a(t)(s - s₀),
where:
v = final velocity (when the ball reaches its maximum height),
s = current height above the initial height.
Since the velocity becomes zero at the maximum height, v = 0. Therefore, the equation simplifies to:
0 = 79² + 2(-32)(s - 4).
Now, we can solve for the maximum height (s):
0 = 6241 - 64(s - 4),
64(s - 4) = 6241,
64s - 256 = 6241,
64s = 6497,
s = 6497/64.
The maximum height the ball will reach is approximately s = 101.52 ft (rounded to two decimal places).