A ball is thrown straight up and passes point B(at a height of 45.6m above its starting point O) in 4.9s

What was its initial speed? The acceleration of gravity is 9.8m/s^2

Answer in units of m/s.

I think the answer is -14.7m/s

but im not sure since you probably cant have negative speed

To find the initial speed of the ball, we can use the kinematic equation:

h = v_0t + (1/2)gt^2

Where:
h is the height (45.6m)
v_0 is the initial velocity (what we need to find)
t is the time taken (4.9s)
g is the acceleration due to gravity (-9.8m/s^2)

Rearranging the equation, we have:

v_0 = (h - (1/2)gt^2) / t

Substituting the given values, we get:

v_0 = (45.6 - (1/2)(-9.8)(4.9^2)) / 4.9

v_0 = (45.6 + 119.828) / 4.9

v_0 = 165.428 / 4.9

v_0 ≈ 33.784 m/s

Therefore, the initial speed of the ball is approximately 33.784 m/s. Note that speed is a scalar quantity and can be positive or negative, but in this case, we only consider the magnitude of the speed.