A ball is thrown vertically upwards at 4ms-1 and returns yo the throwers hand.calculate the maximum height reached by the ball
its initial kinetic energy is converted to gravitational potential energy, so at the top,
finalPE=initial KE
mgh=1/2 m v^2
h=1/2 4^2/9.8=8/9.8 meters.
To calculate the maximum height reached by the ball, we need to apply the equations of motion for vertical motion.
Step 1: Identify the given information:
- Initial velocity (u) = 4 m/s (upwards)
- Final velocity (v) = -4 m/s (downwards, as the ball returns to the thrower's hand)
- Acceleration (a) = -9.8 m/s² (since the force of gravity acts downward)
Step 2: Determine the time it takes for the ball to reach the highest point.
We can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
- Rearranging the equation, we have: t = (v - u) / a
Substituting the values:
t = (-4 m/s - 4 m/s) / -9.8 m/s²
t = -8 m/s / -9.8 m/s²
t ≈ 0.82 seconds (rounded to two decimal places)
Step 3: Calculate the maximum height (h) using the formula:
h = u * t + (1/2) * a * t²
Substituting the values:
h = 4 m/s * 0.82 s + (1/2) * (-9.8 m/s²) * (0.82 s)²
h ≈ 3.28 m (rounded to two decimal places)
Therefore, the maximum height reached by the ball is approximately 3.28 meters.