As a protest against the umpire’s calls, a base- ball pitcher throws a ball straight up into the air at a speed of 20 m/s. In the process, he moves his hand through a distance of 1.47 m.
The acceleration of gravity is 9.8 m/s2 .
If the ball has a mass of 0.18 kg, find the force he exerts on the ball to give it this upward speed.
Answer in units of N.
F = m g + m a
average speed in hand = (20 + 0)/2 = 10 m/s
so
time in hand = 1.47 m/10 m/s = .147 seconds
a = change in speed / time in hand = 20/.147
so F = .18(9.8 + 20/.147)
WOW thx
To find the force the pitcher exerts on the ball, we can use Newton's second law, which states that force is equal to mass multiplied by acceleration.
First, let's find the acceleration of the ball when thrown straight up. Since the ball is moving against the force of gravity, the acceleration will be negative. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Given:
Initial velocity (u) = 20 m/s
Distance (s) = 1.47 m
Acceleration due to gravity (a) = -9.8 m/s^2 (negative sign indicates upward direction)
Rearranging the equation, we get:
v^2 = u^2 + 2as
(0 m/s)^2 = (20 m/s)^2 + 2(-9.8 m/s^2)(1.47 m)
Simplifying the equation:
0 = 400 m^2/s^2 - 29.016 m^2/s^2
0 = 370.984 m^2/s^2
Therefore, the final velocity of the ball is 0 m/s when it reaches its maximum height.
Using Newton's second law, we can find the force:
Force (F) = mass (m) × acceleration (a)
Given:
Mass (m) = 0.18 kg
Acceleration (a) = -9.8 m/s^2 (negative sign indicates upward direction)
Substituting the values into the equation:
F = 0.18 kg × (-9.8 m/s^2)
Calculating:
F = -1.764 N
The force exerted by the pitcher on the ball to give it this upward speed is approximately -1.764 N (negative sign indicates upward direction).
To find the force exerted on the ball, we need to use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a), or F = ma.
In this case, we need to find the acceleration of the ball. Since the ball is thrown straight up into the air, the only force acting on it is gravity. The acceleration due to gravity is always directed downward with a value of -9.8 m/s^2 (negative because it opposes the upward motion of the ball).
We are given the initial velocity of the ball as it is thrown, which is 20 m/s straight up. Since the ball eventually comes to a stop at its highest point, we can use one of the kinematic equations to find the time it takes to reach that point. The kinematic equation we can use is:
v_f = v_i + at,
where v_f is the final velocity (0 m/s at the highest point), v_i is the initial velocity (20 m/s), a is the acceleration (-9.8 m/s^2), and t is the time.
Plugging in the known values, we get:
0 = 20 + (-9.8)t.
Solving for t, we find:
t = -20 / (-9.8) = 2.04 s.
Now that we know how long it takes for the ball to reach its highest point, we can calculate the distance it travels during this time. Using the displacement equation:
s = v_i * t + 1/2 * a * t^2,
where s is the displacement (1.47 m), v_i is the initial velocity (20 m/s), a is the acceleration (-9.8 m/s^2), and t is the time (2.04 s), we can solve for the displacement.
Plugging in the known values, we get:
1.47 = 20 * 2.04 + 1/2 * (-9.8) * (2.04)^2.
Simplifying, we find:
1.47 = 40.8 - 20.2.
Now, let's solve for the acceleration of the ball using the equation:
s = v_i * t + 1/2 * a * t^2.
Rearranging the equation to solve for a, we get:
a = 2 * (s - v_i * t) / t^2.
Plugging in the known values, we get:
a = 2 * (1.47 - 20 * 2.04) / (2.04)^2.
Simplifying, we find:
a = -19.85 m/s^2.
Finally, we can calculate the force exerted on the ball using Newton's second law:
F = m * a.
Plugging in the mass of the ball (0.18 kg) and the acceleration (-19.85 m/s^2), we get:
F = 0.18 * (-19.85) = -3.57 N.
Since force is a vector quantity and the question specifically asks for the force in units of N (Newtons), we can ignore the negative sign and the magnitude of the force exerted on the ball is approximately 3.57 N.