Question

If you throw a 0.5kg ball straight up with an initial speed of 36m/s, how high will it climb? Round to the nearest tenth and include the appropriate unit

Answer 1

Data we have:

m: mass; m = 0.5kg

g: gravitational acceleration; g = -9.81m/s^2 (negative since this acceleration is downward)

vi: initial velocity; vi = 36m/s (positive since this velocity is upward)

vf: final velocity (velocity at the top); vf = 0m/s

We need to solve for d: distance it travels, so let's use this formula:

vf^2 = vi^2 + 2gd

Let's plug in the variables we know:

0^2 = 36^2 + 2*(-9.81)*d

Solve for d:

Move 36^2 to the other side, then negate both sides to remove the negative sign from both sides:

36^2 = 2*9.81*d

Divide by (2*9.81) to isolate d:

d = 36^2/(2*9.81)

Finally, solve:

d = 66.1 m