A ball is tossed vertically upwards with a speed of 5.0 m s–1. After how many seconds will the ball return to its initial position?

plz help !

hf=hi+vi*t+1/2 g t^2

g=-9.8
0=0+5t-4.8t^2

that is a quadratict equation, use the quadratic formula to find t.

or, solve it by factoring.... ;)

V = Vo + g*Tr.

0 = 5 + (-9.8)Tr,
Tr = 0.51 s. = Rise time.

Tf = Tr = 0.51 s. - Fall time.

Tr+Tf = 0.51 + 0.51 = 1.02 s. = Return time.

To find the time it takes for the ball to return to its initial position, you can use the formula for the time of flight of an object in projectile motion.

Since the ball is thrown vertically upwards, you can use the equation:

t = v / g

Where:
t is the time of flight,
v is the initial vertical velocity of the ball, and
g is the acceleration due to gravity.

In this case, the initial vertical velocity (v) is 5.0 m/s, and the acceleration due to gravity (g) is approximately 9.8 m/s².

Substituting the given values into the equation, you can calculate the time it takes for the ball to return to its initial position:

t = 5.0 m/s / 9.8 m/s²

Simplifying this calculation, you will get:

t ≈ 0.51 seconds

Therefore, the ball will return to its initial position after approximately 0.51 seconds.