An elementary student kicks a ball straight into the air with a velocity of 16 feet/sec. If acceleration due to gravity is -32 ft/sec^2, how many seconds after it leaves his foot will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
Are you sure that's how you would do that? Because its not one of my options. my options are 1.25, 6, 0.50 or cannot be determined.
To find the time it takes for the ball to reach its highest point, we need to determine when its velocity becomes zero.
We know that the acceleration due to gravity is always acting downwards and has a value of -32 ft/sec^2. Since the ball is going straight up, the acceleration due to gravity will oppose the motion of the ball, causing its velocity to decrease.
Let's set up an equation to solve for the time when the velocity becomes zero. We know that the initial velocity (v₀) is 16 ft/sec, the acceleration (a) is -32 ft/sec^2, and we want to find the time (t) when the velocity (v) is zero.
The equation we can use is:
v = v₀ + a * t
Setting v = 0, v₀ = 16 ft/sec, and a = -32 ft/sec^2, we get:
0 = 16 - 32t
Rearranging the equation to solve for t:
32t = 16
t = 16/32
t = 0.5 seconds
Therefore, it will take the ball 0.5 seconds to reach its highest point after it leaves the student's foot.
the height h is
h = 16t - 32t^2
as with any parabola, the vertex is at t = -b/2a = -16/-64 = 1/4 second
h = 16t(1-2t)
The roots are 0 and 1/2. The vertex lies halfway between the roots, at t = 1/4.