A ball is thrown upward with an initial velocity of 4ft/sec from a height of ten feet. what is the maximum height that the ball will reach.
h(t)= -16+vt+s
v = Vi - g t
at top v = 0
0 = 4 -32 t
t = 1/8 second
(4 ft /sec is really slow. I bet you mean 40)
h = 10 + 4 (1/8) - 16 (1/64)
= 10 + .5 - .25
= 10.25 ft
To determine the maximum height reached by the ball, we need to find the peak of its trajectory. The equation that describes the height of an object thrown upward is given by h(t) = -16t^2 + vt + s, where h(t) represents the height at time t, v represents the initial velocity, and s represents the initial height.
Given that the ball is thrown upward with an initial velocity of 4 ft/sec from a height of 10 feet, we can plug these values into the equation:
h(t) = -16t^2 + 4t + 10
To find the maximum height, we need to determine the value of t when the ball reaches its peak. At the peak, the velocity of the ball becomes zero.
We can find this time by taking the derivative of h(t) with respect to t and setting it equal to zero:
h'(t) = -32t + 4 = 0
Solving this equation, we get:
-32t = -4
t = -4 / -32
t = 1/8 seconds
To find the maximum height, plug this time value back into the equation:
h(1/8) = -16(1/8)^2 + 4(1/8) + 10
Simplifying, we get:
h(1/8) = -2 + 1/2 + 10
h(1/8) = 8.5 ft
Therefore, the maximum height reached by the ball is 8.5 feet.