The height h is the time of seconds that the ball has been in the air with an initial vertical velocity of 48 feet per second from a height of 5 feet above the ground is given by the equation
h=-16t^2 + 48t + 5
where t is the time in seconds that the ball has been in the air. After how many seconds in the ball at of 37 feet.
To find out after how many seconds the ball is at a height of 37 feet, we can set the height equation equal to 37 and solve for t.
The equation is: h = -16t^2 + 48t + 5
Substituting h = 37, we get: 37 = -16t^2 + 48t + 5
To solve this quadratic equation, we can rearrange it to the form: -16t^2 + 48t + 5 - 37 = 0
Simplifying further: -16t^2 + 48t - 32 = 0
Now, we can either factor this quadratic equation or use the quadratic formula to find the roots.
Let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 48, and c = -32.
Plugging in these values, we get: t = (-48 ± √(48^2 - 4(-16)(-32))) / (2(-16))
Simplifying: t = (-48 ± √(2304 - 2048)) / (-32)
This becomes: t = (-48 ± √256) / (-32)
Simplifying the square root: t = (-48 ± 16) / (-32)
Now, we have two possible values for t:
t1 = (-48 + 16) / (-32) = -32 / (-32) = 1
t2 = (-48 - 16) / (-32) = -64 / (-32) = 2
So, after 1 second and 2 seconds, the ball is at a height of 37 feet.
Therefore, after 1 or 2 seconds, the ball is at a height of 37 feet.
I have not seen a gravity problem in feet instead of meters for a while.
37 = -16 t^2 + 48 t + 5
16 t^2 - 48 t + 32 = 0
t^2 - 3 t + 2 = 0
(t-2)(t-1) = 0
t = 1 or 2
so
the ball is at 37 feet after one second on the way up and at two seconds on the way down