A ball is thrown upward with an initial velocity of 4 ft/sec from a height of 10 feet. What is the maximum height that the ball will reach?
h(t)=-16t^2+v0t+s0
Please help im lost
plug in your numbers:
h(t) = -16t^2 + 4t + 10
Recall from your Algebra I that this is just a parabola, with vertex at x = -b/2a, or in this case, x = 1/8
h(1/8) = 41/4
This is no time to forget your earlier math!
To find the maximum height that the ball will reach, we need to determine when the ball reaches its highest point, which occurs when the vertical velocity is 0.
Given:
Initial velocity, v0 = 4 ft/sec (upward)
Initial height, s0 = 10 ft
Acceleration due to gravity, a = -32 ft/sec^2 (negative because gravity acts downward)
Using the equation for vertical motion:
h(t) = -16t^2 + v0t + s0
To find the time when the ball reaches its maximum height, we set the velocity component equal to 0:
v(t) = v0 + at = 0
4 - 32t = 0
Solve for t:
32t = 4
t = 4/32
t = 1/8 sec
Substitute this value of t back into the equation for height to find the maximum height:
h(t) = -16(1/8)^2 + 4(1/8) + 10
Simplifying:
h(t) = -16(1/64) + 4/8 + 10
h(t) = -1/4 + 1/2 + 10
h(t) = -1/4 + 2/4 + 40/4
h(t) = 41/4
h(t) = 10.25 ft
Therefore, the maximum height that the ball will reach is 10.25 feet.
To find the maximum height that the ball will reach, we can use the quadratic equation h(t) = -16t^2 + v0t + s0, where h(t) represents the height of the ball at time t, v0 represents the initial velocity, and s0 represents the initial height.
Given: v0 = 4 ft/sec (initial velocity) and s0 = 10 ft (initial height)
To find the maximum height, we need to determine the time that corresponds to the maximum height.
Since the ball is thrown upward, it will reach its maximum height when its velocity becomes zero. At that point, the ball will start to fall back down.
To find the time when the velocity becomes zero, we can use the formula for velocity:
v(t) = v0 - 32t
Set v(t) = 0 and solve for t:
0 = v0 - 32t
32t = v0
32t = 4
t = 4/32
t = 1/8
Now that we have the time at which the velocity becomes zero (t = 1/8 sec), we can substitute this value back into the height equation to find the maximum height.
h(t) = -16t^2 + v0t + s0
h(1/8) = -16 * (1/8)^2 + 4 * (1/8) + 10
h(1/8) = -16 * (1/64) + 4/8 + 10
h(1/8) = -1/4 + 1/2 + 10
h(1/8) = -1/4 + 2/4 + 40/4
h(1/8) = (39/4) feet
Therefore, the maximum height that the ball will reach is 39/4 feet or approximately 9.75 feet.