Question on vertical motion - gravity?

Question

Throw a ball upward from point O with an

initial speed of 74 m/s.

The acceleration of gravity is 9.8 m/s2 .

What is the maximum height ? Answer

in units of m.

If the speed of the ball as it passes point B is

(74 m/s) = 37 m/s, what is the height

of B above O? Answer in units of m.

Answer 1

The initial speed of the ball is Vi = 74 m/s

The final speed at the Max height point should be zero. So we will set Vf (final) = 0 m/s

Use the equation: (Vf)^2 = (Vi)^2 + 2AX

A (acceleration) = - 9.8 m/s2 (negative because it is in the opposite direction of velocity - we set the upwards direction positive)

X (displacement) - That's what we want to find

Plug in:

0=74^2 - 2 * 9.8 * X

19.6X = 74^2

X=279.38 m

Now for the second part:

If when it passes point B it has velocity that is smaller than 74, then it must be above point O (because velocity decreases with time)

use the same equation, but now plug in the new velocity as Vf and it will give you the distance the ball traveled from O to B.

(37)^2 = (74)^2 - 2 * A * X

19.6X = (74)^2 - (37)^2

19.6X = 4107

X=209.54 m //That's the height of B above O

I hope it was helpful

Answer 2

74/9.8=7.55

7.55*74/2=280.105m is the maximom height

74-37=37

37/9.8=3.77

3.77*(37+74)/2=209.235 is the height of B above o