Question

If you throw a ball up into the air and it reaches its original height after 6 seconds, calculate how high above its original position the ball reached. (Answer to two decimal places).

Answer 1

Explanation:

Assuming that there is no air resistance, the given time means that the ball reached its maximum height 3 seconds after throwing it upwards so we can solve its

maximum height by solving for $v_{0y}$ first:

$v_y = v_{0y} - gt \Rightarrow v_{0y} = gt$

$v_{0y} = (9.8\:\text{m/s}^2)(3\:\text{s}) = 29.4\:\text{m/s}$

Now that we know $v_{0y},$ we can now solve for $y_{max}:$

$y_{max} = v_{0y}t - \frac{1}{2}gt^2$

$\:\:\:\:\:\:=(29.4\:\text{m/s})(3\:\text{s}) - \frac{1}{2}(9.8\:\text{m/s}^2)(3\:\text{s})^2$

$\:\:\:\:\:\:=44.1\:\text{m}$