A boy tosses a coin upward with a velocity of 20m/s

What will be the maximum height?
Time of flight?
Velocity as it returns to the hand?
Velocity as it hits the ground?

v = Vi + a t

so at top
0 = 20 - 9.81 t
solve for t at top
T = time in air = 2 t

h = h of hand + 20 t - 4.9t^2
solve for h

v at hand = -20 by symmetry

I do not know how high his hand was so can say nothing about ground

To calculate the maximum height, time of flight, velocity as it returns to the hand, and velocity as it hits the ground, we can use the equations of motion for projectile motion.

1. Maximum Height:
The coin is thrown upwards, so its initial velocity (u) is 20 m/s. The final velocity (v) at the maximum height is 0 m/s because the coin momentarily stops there. Acceleration due to gravity (g) is approximately 9.8 m/s^2. We can use the equation:

v^2 = u^2 + 2gh

Since v = 0 at the maximum height, we can solve for h (the maximum height):

0^2 = (20 m/s)^2 - 2 * 9.8 m/s^2 * h

400 m^2/s^2 = 19.6 m/s^2 * h

h = 400 m^2/s^2 / 19.6 m/s^2

h ≈ 20.4 meters

Therefore, the maximum height is approximately 20.4 meters.

2. Time of Flight:
The time of flight (t) is the total time it takes for the coin to reach its initial position. We can use the equation:

v = u + gt

Since the initial velocity (u) is 20 m/s and the final velocity (v) is 0 m/s (when the coin returns to its initial position), we can solve for t:

0 m/s = 20 m/s + 9.8 m/s^2 * t

-20 m/s = 9.8 m/s^2 * t

t = -20 m/s / 9.8 m/s^2

t ≈ 2.04 seconds

Therefore, the time of flight is approximately 2.04 seconds.

3. Velocity as it Returns to the Hand:
The velocity as the coin returns to the hand is the same as its initial velocity since there is no external force acting on it. Therefore, the velocity is 20 m/s.

4. Velocity as it Hits the Ground:
To find the velocity just before it hits the ground, we can use the equation:

v = u + gt

Since the initial velocity (u) is 20 m/s and acceleration due to gravity (g) is -9.8 m/s^2 (negative because it acts in the opposite direction of the initial velocity), we can solve for v:

v = 20 m/s + (-9.8 m/s^2) * t

Substituting the time of flight (t = 2.04 seconds):

v = 20 m/s - 9.8 m/s^2 * 2.04 s

v ≈ 0 m/s

Therefore, the velocity just before the coin hits the ground is approximately 0 m/s (which means it momentarily comes to rest).