if a ball is thrown into the air at a speed of 22 m/s what is its position after 2.8 seconds?

Assuming it is thrown straight up, the altitude y after t seconds is

y = 22t - (g/2)t^2

where g = 9.8 m/s^2 is the acceleration of gravity

I get 23.2 meters

Thank you so much!

To determine the position of the ball after 2.8 seconds, we can use the equations of motion. In this case, we need to use the equation for free fall:

s = ut + (1/2)at²

where:
s: position or displacement
u: initial velocity
t: time
a: acceleration due to gravity

In this scenario, since the ball is thrown upwards, the value of acceleration due to gravity (a) will be negative (-9.8 m/s²), as it acts in the opposite direction of the motion. The initial velocity (u) will be +22 m/s (positive because it is directed upwards) and the time (t) is given as 2.8 seconds.

Plugging these values into the equation, we get:

s = (22 m/s)(2.8 s) + (1/2)(-9.8 m/s²)(2.8 s)²

simplifying:

s = 61.6 m + (-38.864 m)

s = 22.736 m

Therefore, the position of the ball after 2.8 seconds will be approximately 22.736 meters above its initial position.