Answer:
(a) 1.28 m
(b) 0.51 s
Explanation:
Given:
Initial speed, U = 5 m/s
Consideration;
1. When the coin is tossed upward, the coin is moving against gravity. Therefor, the acceleration due to gravity experienced by the coin is negative, -g = -9.8 m/s².
2. The final velocity of the coin in the air is zero.
From equation of motion under free fall,
V² = U² + 2gh
h = [tex]h = \frac{V^{2} - U^{2}}{2(-g)}[/tex]
Where;
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
h is the height
[tex]h = \frac{0^{2} - 5^{2}}{2(-9.8)}[/tex]
[tex]h = \frac{0^{-25}{19.60}[/tex]
h = 1.2755 m
h = 1.28 m
(b) The coin still has its initial velocity, u to be equal to 5 m/s, final velocity, v to be equal to 0 m/s and acceleration due to gravity to be -9.8 m/s while in the air before returning to the release point.
From equation of motion under free fall,
V = U + gt
[tex]t = \frac{V - U}{g}[/tex]
[tex]t = \frac{0 - 5}{-9.8}[/tex]
t = 0.51 s