A fast, measured pitched baseball left the pitcher's hand at a speed of 49.0 m/s. The pitcher was in contact with the ball over a distance of 1.59 m and produced constant acceleration.
(a) What acceleration did he give the ball?
accelearation=changevelocty/time
oops, too quick
Vf^2=2ad solve for a.
To find the acceleration given to the ball, you need to use the formula that relates acceleration, initial velocity, final velocity, and distance.
The formula is:
(v^2 - u^2) = 2ad
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
d is the distance covered.
In this case, the initial velocity (u) is 0 since the ball starts from rest in the pitcher's hand. The final velocity (v) is the speed at which the ball leaves the pitcher's hand, given as 49.0 m/s. The distance (d) over which the pitcher is in contact with the ball is 1.59 m.
Plugging in the values into the formula:
(49^2 - 0^2) = 2a(1.59)
Simplifying:
2401 = 3.18a
To find the acceleration (a), divide both sides of the equation by 3.18:
a = 2401 / 3.18
a ≈ 755.97 m/s^2
So, the acceleration given to the ball by the pitcher is approximately 755.97 m/s^2.