Question

Suppose that a person throws a ball upward at 15 m/s while standing on the edge of a cliff so that the ball can fall to the base of the cliff 50 m below.

a) How long does it take the ball to reach the base of the cliff?

b) What is the total distance covered by the ball.

Answer 1

a) We can calculate the time with the equation

$h= h_0+V_0t-\frac{gt^2}{2}= 0 \\ t= \frac{-V_0 ±\sqrt{V_0^2+2gh_0 }}{-g} \\ t= \frac{V_0}{g} \mp \sqrt{ (\frac{V_0}{g})^2+\frac{2h_0}{g} } \\ t= (\frac{V_0}{g})\Big(1\mp \sqrt{ 1+\frac{2gh_0}{V_0^2} } \Big)$

We take the positive root and we calculate the total time with V₀ = 15 m/s and h₀ = 50 m:

$t= (\frac{15\,m/s}{9.8\,m/s^2})\Big( \sqrt{ 1+\frac{2(9.8\,m/s^2)(50\,m)}{(15\,m/s)^2} } +1\Big) \\ t= ( 1.53\,s)(1+\sqrt{ 1+4.355 })= 5.07\,s$

b) the total distance along the y axis will be twice the maximum distance reached when the ball goes upwards and then we add it to the height of the cliff h₀ to find the total distance traveled by the ball as D:

$t_{upwards}= \frac{V_0}{g} \\ d=V_0t-\frac{gt^2}{2}=V_0(\frac{V_0}{g})-\frac{g}{2}(\frac{V_0}{g})^2 \\ \therefore d = \frac{V_0^2}{g}-\frac{V_0^2}{2g}=\frac{V_0^2}{2g} \\ \to D = h_0 + 2d = h_0 +\frac{V_0^2}{g} \\ D= 50\,m+\frac{[15\frac{m}{s}]^2}{9.8\frac{m}{s^2}}= 72.96\,m$

In conclusion, a) it takes 5.07 s for the ball to reach the base of the cliff, and b) the ball covers a total distance of 72.96 m.