Asked by jason
An outfielder throws a 1.88 kg baseball at a
speed of 57 m/s and an initial angle of 10.2.
What is the kinetic energy of the ball at the
highest point of its motion?
Answer in units of J
speed of 57 m/s and an initial angle of 10.2.
What is the kinetic energy of the ball at the
highest point of its motion?
Answer in units of J
Answers
Answered by
Sabrina
To find the KE of the ball at the highest ball of motion, we have to use the initial velocity and angle to find the horizontal velocity.
V(horizontal) = V(initial) * cos(theta)
V(horizontal) = (57 m/s) * cos(10.2)
V(horizontal) = 56.099 m/s
Using the velocity in the horizontal direction that we found, along with the mass given, we can calculate the KE of the ball at its highest point.
KE = (1/2)mv^2
KE = (1/2)(1.88 kg)(56.099 m/s)^2
KE = 2958.27 J
V(horizontal) = V(initial) * cos(theta)
V(horizontal) = (57 m/s) * cos(10.2)
V(horizontal) = 56.099 m/s
Using the velocity in the horizontal direction that we found, along with the mass given, we can calculate the KE of the ball at its highest point.
KE = (1/2)mv^2
KE = (1/2)(1.88 kg)(56.099 m/s)^2
KE = 2958.27 J
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