A ball of mass 8 kg is thrown vertically upward from the ground , with an initially velocity of 25 m/s. Determine the maximum height to which it will travel if atmospheric resistance is neglected.
h(t) = 25t - 4.9t^2 = t(25-4.9t)
max height is reached halfway between 0 and 25/4.9, at t=2.55
h(2.55) = 31.89
ΔKE= - ΔPE
KE2-KE1 =-(PE2 -PE1)
0 - m•v²/2 =0 – m•g•h.
h= v²/2g=25²/2•9.8 = 31.89 m
Please help me to elaborate the answer as i tried but couldn't
A ball of mass 8 kg is thrown vertically upward from the ground , with an initially velocity of 25 m/s. Determine the maximum height to which it will travel if atmospheric resistance is neglected
To determine the maximum height to which the ball will travel, we can use the equations of motion. Let's break down the problem step by step.
Step 1: Identify the given values:
- Mass (m) of the ball: 8 kg
- Initial velocity (u) of the ball: 25 m/s
- Acceleration due to gravity (g): 9.8 m/s^2 (considered positive for upward motion)
Step 2: Determine the final velocity (v) at the maximum height:
At the maximum height, the final velocity becomes zero (v = 0) since the ball will momentarily stop before falling back down.
Step 3: Apply the equation of motion:
The equation of motion relating initial velocity (u), final velocity (v), acceleration (a), and displacement (s) is:
v^2 = u^2 + 2as
At the maximum height, v = 0, so the equation becomes:
0 = u^2 + 2as
Step 4: Solve for the displacement (s):
Rearrange the equation to solve for s:
s = -u^2 / (2a)
Step 5: Substitute the given values and calculate:
s = - (25 m/s)^2 / (2 * (-9.8 m/s^2))
s = -625 m^2/s^2 / (-19.6 m/s^2)
s ≈ 31.89 m (rounded to two decimal places)
Therefore, the maximum height to which the ball will travel is approximately 31.89 meters when neglecting atmospheric resistance.