A ball of mass 400g is thrown with an initial velocity of 30.0 ms-1 at an angle of 45.0 degrees to the horizontal.air resistance is negligible.the ball reaches a maximum height H after a time of 2.16s (1)calculate the initail kinetic energy of the ball? (2) the maximum height H of the ball?
Maximum height is 45.918m and the initial kinetic energy is 180J
To calculate the initial kinetic energy of the ball, we can use the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Given:
Mass of the ball = 400g = 0.4kg
Initial velocity = 30.0 m/s
Substituting these values into the formula, we get:
Kinetic Energy = (1/2) * 0.4kg * (30.0 m/s)^2
Now, let's calculate the value:
Kinetic Energy = (1/2) * 0.4kg * (900.0 m^2/s^2)
= 180.0 Joules
Therefore, the initial kinetic energy of the ball is 180.0 Joules.
Now, to calculate the maximum height (H) of the ball, we can use the following equations of motion:
Vertical Displacement (h) = Vertical Initial Velocity (Viy) * Time (t) - (1/2) * Acceleration due to Gravity (g) * Time (t)^2
At the maximum height, the ball's vertical velocity (Viy) becomes zero, and the acceleration due to gravity (g) is acting in the downward direction. So, we can write:
0 = Viy * t - (1/2) * g * t^2
Substituting the given values:
0 = (30.0 m/s) * 2.16s - (1/2) * (9.8 m/s^2) * (2.16s)^2
Now, let's calculate the value:
0 = 64.8m - 22.6304m
= 42.1696m
Therefore, the maximum height (H) of the ball is 42.17 meters.
Physics sorry
Use
sum of potential and kinetic energies = constant
when air resistance is neglibible.
Potential energy = mgh at a height h from datum
kinetic energy = (1/2)mv^2 when moving at a speed of v.