Kinematics: Question 10

Worked solution

Setting up a sign convention

Take downwards as positive throughout, since both the initial velocity and the acceleration due to gravity act downwards for this stone. So $u=6.00\text{ m s}^{-1}$, $a=g=+9.81\text{ m s}^{-2}$, and the displacement to the base of the cliff is $s=45.0\text{ m}$.

Part (a): Time for the thrown stone to reach the base

Using $s = ut + \tfrac{1}{2}at^2$ substitute the values and rearrange into the standard quadratic form $At^2+Bt+C=0$: $45.0 = 6.00t + \tfrac{1}{2}(9.81)t^2$ $4.905t^2 + 6.00t - 45.0 = 0$

Applying the quadratic formula $t=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}$ with $A=4.905$, $B=6.00$, $C=-45.0$: $B^2-4AC = (6.00)^2 - 4(4.905)(-45.0) = 36.0 + 882.9 = 918.9$ $t = \frac{-6.00 \pm \sqrt{918.9}}{2(4.905)} = \frac{-6.00 \pm 30.31}{9.81}$

The negative root gives a negative time, which is not physically meaningful, so: $t_1 = \frac{-6.00 + 30.31}{9.81} = \frac{24.31}{9.81}$ $t_1 = 2.48\text{ s} (3\text{ s.f.})$

Part (b): Speed on hitting the base of the cliff

Using $v = u + at_1 = 6.00 + (9.81)(2.478)$ $v = 6.00 + 24.31 = 30.3\text{ m s}^{-1} (3\text{ s.f.})$

Check: using $v^2=u^2+2as=(6.00)^2+2(9.81)(45.0)=36.0+882.9=918.9$, so $v=\sqrt{918.9}=30.3\text{ m s}^{-1}$. This matches, and is the same square root that appeared in the quadratic formula in part (a), since both come from the same underlying relationship.

Part (c): Comparing with a stone dropped from rest

For the second stone, $u=0$, so the equation of motion simplifies to $s=\tfrac{1}{2}gt^2$: $45.0 = \tfrac{1}{2}(9.81)t_2^2$ $t_2^2 = \frac{2(45.0)}{9.81} = \frac{90.0}{9.81} = 9.174$ $t_2 = \sqrt{9.174} = 3.03\text{ s} (3\text{ s.f.})$

The difference between the two fall times is: $\Delta t = t_2 - t_1 = 3.029 - 2.478$ $\Delta t = 0.550\text{ s} (3\text{ s.f.})$

This makes physical sense: the thrown stone has a “head start” in speed, so it reaches the ground sooner than the dropped stone, even though both experience exactly the same acceleration $g$ throughout the fall.

Final answers

• (a) Time for the thrown stone to fall $= \boxed{2.48}\text{ s}$
• (b) Speed on landing $= \boxed{30.3}\text{ m s}^{-1}$
• (c) Time for the dropped stone $= \boxed{3.03}\text{ s}$; difference in fall times $= \boxed{0.550}\text{ s}$