A projectile is launched upward from a height of 960 feet with an initial velocity of 64 ft/s. The equation gives the height h after t seconds. Find the number of seconds until it returns to the ground.
just solve the equation when y=0. Just a normal quadratic solution.
To find the number of seconds until the projectile returns to the ground, we need to determine when the height of the projectile is zero.
The equation that gives the height h of the projectile after t seconds is:
h(t) = -16t^2 + vt + h
Where:
-16 is the gravitational constant (acceleration due to gravity in the US customary unit system)
v is the initial velocity (64 ft/s)
h is the initial height (960 ft)
Setting h(t) = 0, we can solve for t to find when the height is zero.
0 = -16t^2 + 64t + 960
We have a quadratic equation, so we can use the quadratic formula to find the values of t that satisfy the equation.
The quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -16, b = 64, and c = 960.
Let's plug these values into the quadratic formula to find the solutions for t:
t = (-64 ± √(64^2 - 4*(-16)*960)) / (2*(-16))
Simplifying further:
t = (-64 ± √(4096 + 61440)) / (-32)
t = (-64 ± √65536) / (-32)
t = (-64 ± 256) / (-32)
This gives two possible values for t:
t1 = (-64 + 256) / (-32) = 192 / (-32) = -6
t2 = (-64 - 256) / (-32) = -320 / (-32) = 10
Since time cannot be negative, we ignore t1 and consider t2. Therefore, the number of seconds until the projectile returns to the ground is 10 seconds.