If a rocket is shot upward with an initial speed of 176 feet per second from a platform 64 feet above the ground, when will the rocket be at its maximum height and how high will it be?
Hint: The graph of a parabola is symmetric.
To find when the rocket will be at its maximum height and how high it will be, we need to consider the motion of the rocket using basic kinematic equations.
Let's assume the upward direction is positive. Given that the rocket is shot upward with an initial speed of 176 feet per second, we can use the equation for vertical motion:
y = y0 + v0*t - (1/2)g*t^2
Where:
y - final height of the rocket
y0 - initial height of the rocket
v0 - initial velocity of the rocket
g - acceleration due to gravity (approximately 32 feet per second squared)
t - time
Using the given values:
y0 = 64 feet
v0 = 176 feet per second
g = 32 feet per second squared
We need to find the time when the rocket reaches its maximum height. At the maximum height, the velocity of the rocket becomes zero. So, we can set v = 0 and solve for t:
0 = v0 - g*t
g*t = v0
t = v0 / g
Now, substitute this value of t back into the equation to find the maximum height:
y = y0 + v0*t - (1/2)g*t^2
Plug in the values:
y = 64 + 176*(v0 / g) - (1/2)*g*(v0 / g)^2
Simplifying:
y = 64 + 176*(v0 / g) - (1/2)*(v0^2 / g)
Calculating the values:
t = 176 / 32 = 5.5 seconds
y = 64 + 176*(5.5) - (1/2)*(176^2 / 32)
y = 64 + 880 - 308
y = 64 + 572
y = 636 feet
Therefore, the rocket will be at its maximum height after 5.5 seconds, and it will reach a height of 636 feet.