A ball is thrown straight up from the top of a hill 30 feet high with an initial velocity of 72ft/sec. How high above level ground will the ball get? (Objects subjected to gravity adhere to s(t) = -16t^2 + v0t + so where s is the height of the object in feet v0 is the initial velocity and s0 is the initial height)
Your equation will be
s(t) = -16t^2 + 72t + 30
We need the vertex of this parabola, the t of the vertex is -b/2a
or -72/-32 = 2.25 seconds
s(2.25) = -16(2.25)^2 + 72(2.25) + 30
= ..... m
Well, if the ball is thrown straight up from the top of a hill, it's really asking for trouble. I mean, why would you throw a ball off a hill instead of playing catch with someone? It's just creating unnecessary problems!
But, getting back to the question, we can use the formula s(t) = -16t^2 + v0t + s0 to find the height of the ball at any given time.
In this case, the initial height (s0) is 30 feet, and the initial velocity (v0) is 72 ft/sec. So we can plug those values into our formula:
s(t) = -16t^2 + 72t + 30
Now, to find how high the ball will get, we need to find the time at which the ball reaches its highest point. At the highest point, the velocity of the ball is 0, so we can find the time by setting the derivative of s(t) equal to 0:
s'(t) = -32t + 72
0 = -32t + 72
Solving this equation, we get t = 2.25 seconds.
Now we can find the height of the ball at this time by plugging t = 2.25 into our formula:
s(2.25) = -16(2.25)^2 + 72(2.25) + 30
After some calculations, we find that the ball reaches a maximum height of approximately 128.6 feet above level ground.
So, the ball gets as high as a kangaroo on stilts trying to reach a tree branch. Quite impressive!
To find the maximum height reached by the ball, we need to determine the time it takes for the ball to reach its highest point. At this point, the vertical velocity of the ball will be zero.
Given:
Initial velocity v0 = 72 ft/sec
Initial height s0 = 30 feet
Acceleration due to gravity a = -32 ft/sec^2 (negative because it acts in the opposite direction to the motion)
Using the equation s(t) = -16t^2 + v0t + s0, we can set the vertical velocity equal to zero and solve for time (t):
0 = -16t^2 + 72t + 30
This equation can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 72, and c = 30.
t = (-72 ± √(72^2 - 4(-16)(30))) / (2(-16))
Simplifying further:
t = (-72 ± √(5184 + 1920)) / -32
t = (-72 ± √7104) / -32
t = (-72 ± 84.3) / -32 (taking the positive value because time cannot be negative)
t = (12.3) / -32 = -0.384
Since time cannot be negative, we ignore this negative value.
Therefore, the time it takes for the ball to reach its highest point is approximately 0.384 seconds.
To find the maximum height above ground, we substitute this time value back into the equation:
s(t) = -16t^2 + v0t + s0
s(0.384) = -16(0.384)^2 + 72(0.384) + 30
s(0.384) = -2.4576 + 27.648 + 30
s(0.384) = 55.1904
Therefore, the ball will reach a maximum height of approximately 55.19 feet above ground level.
To find the maximum height above level ground that the ball reaches, we can use the given equation for the height of an object subjected to gravity:
s(t) = -16t^2 + v0t + s0
In this case, the initial height (s0) is 30 feet and the initial velocity (v0) is 72 ft/sec. We want to find the maximum height, which occurs when the velocity is zero. So, we need to find the time (t) at which the velocity is zero.
To do this, we set the velocity equation, which is the derivative of the height equation, to zero:
v(t) = s'(t) = -32t + 72 = 0
Solving this equation, we get:
-32t + 72 = 0
-32t = -72
t = 72/32 = 2.25 seconds
Now that we have the time at which the velocity is zero, we can substitute this value back into the height equation to find the maximum height above level ground:
s(2.25) = -16(2.25)^2 + 72(2.25) + 30
Evaluating this equation, we get:
s(2.25) = -16(5.06) + 162 + 30
s(2.25) = -80.96 + 162 + 30
s(2.25) = 111.04 feet
Therefore, the ball will reach a height of approximately 111.04 feet above level ground.