A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is -32 ft/sec2, how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
height = -16t^2 + 72t + 0
so you want the vertex of this parabola
t of the vertex = -72/-32
= 2.25 seconds to reach the max
height = -16(2.25)^2 + 72(2.25) = 81
To find the time it takes for the ball to reach its highest point, we need to determine when the velocity of the ball becomes zero.
We can use the equation of motion:
Vf = Vi + at
Where:
Vf = Final velocity (0 ft/sec, as the ball reaches its highest point)
Vi = Initial velocity (72 ft/sec)
a = Acceleration due to gravity (-32 ft/sec^2)
t = Time
Substituting the given values into the equation, we have:
0 = 72 - 32t
Rearranging the equation to solve for t:
32t = 72
t = 72 / 32
t = 2.25 seconds
Therefore, it will take approximately 2.25 seconds for the baseball to reach its highest point after leaving the pitcher's hand.
To determine the time it takes for the baseball to reach its highest point, we need to first understand the relationship between velocity, acceleration, and time.
The equation that relates velocity, acceleration, and time for an object in free fall near the Earth's surface is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the ball is thrown straight up, so its initial velocity is +72 ft/sec (upward direction), and the acceleration is -32 ft/sec^2 (downward direction due to gravity). The final velocity when the ball reaches its highest point will be 0 ft/sec.
Plugging the values into the equation, we have:
0 = 72 - 32t
To solve for t, we can rearrange the equation:
32t = 72
Dividing both sides by 32:
t = 72 / 32
Simplifying:
t = 2.25 seconds
Therefore, it will take the ball 2.25 seconds to reach its highest point after leaving the pitcher's hand.