Question

A long jumper leaves the ground at$45°$above the horizontal plane and lands 8.0 m away. What is her ‘takeoff’ speed$v_o$? (b) Now she is out on a kike and comes to the left bank of the river. There is no bridge, and the right bank is 10.0 m away horizontally and 2.5 m below vertically. If she long jumps from the edge of the left bank at$45°$with the speed calculated, (a) how long or short of the opposite bank will she land?

Short answer

The jumper's takeoff speed is $8.85\frac{\mathrm{m}}{\mathrm{s}}$.

The jumper lands just on the opposite bank of the river, neither long nor short.

Step-by-step solution

Step 1. Given data

The motion of the jumper can be considered as the motion of a projectile.The jumper's horizontal velocity is the same during the motion, but she travels more horizontal distance when she jumps from a high cliff.

Given data:

The jumper's takeoff speed is $v_o$.

The angle of her projectile motion is $\theta =45°$.

For the first case, she moves a horizontal distance $R_o=8.0\mathrm{m}$.

Assumptions:

Let the jumper take t time to land on the ground when she jumps from the edge of the left bank.

Step 2. Finding the takeoff speed

Part (a)

The range of a projectile motion is

$R=\frac{v_o^2\sin 2\theta }{g}$.

Now, for the first part of the motion,

$\begin{array}{c}{R}_o=\frac{v_o^2\sin 2\theta }{g}\\ 8.0\mathrm{m}=\frac{v_o^2\sin \left(2\times 45°\right)}{9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}}\\ v_o^2=\frac{\left(8.0\mathrm{m}\right)\times \left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}{\sin 90°}\\ v_o=8.85\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the jumper's takeoff speed is $8.85\frac{\mathrm{m}}{\mathrm{s}}$.

Step 3. Calculation of the horizontal distance for the second part

Part (b)

For the vertical motion,

$\begin{array}{c}y-y_o=\left(v_o\sin 45°\right)t-\frac{1}{2}g{t}^2\\ -2.5\mathrm{m}=\left(8.85\frac{\mathrm{m}}{\mathrm{s}}\times \sin 45°\right)t-\left(\frac{1}{2}\times 9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)t^2\\ -2.5\mathrm{m}=6.26 \frac{\mathrm{m}}{\mathrm{s}}t-4.90\frac{\mathrm{m}}{{\mathrm{s}}^2}{t}^2\\ 4.90t^2-\left(6.26\mathrm{s}\right)t-2.5{\mathrm{s}}^2=0\end{array}$

Now, calculating the value of t,

$\begin{array}{c}t=\frac{-\left(-6.26\mathrm{s}\right)\pm \sqrt{{\left(-6.26\mathrm{s}\right)}^2-\left(4\times 4.90\times \left(-2.5{\mathrm{s}}^2\right)\right)}}{2\times 4.90}\\ =\frac{6.26\mathrm{s}\pm \sqrt{{\left(-6.26\mathrm{s}\right)}^2+\left(49{\mathrm{s}}^2\right)}}{9.80}\\ =-0.32\mathrm{s},1.60\mathrm{s}\end{array}$

Here, you can find two values of the time but only $t=1.60\mathrm{s}$is acceptable.

Therefore, the horizontal distance covered by the jumper is

$\begin{array}{c}x=\left(v_o\cos 45°\right)t\\ =\left(8.85\frac{\mathrm{m}}{\mathrm{s}}\times \cos 45°\right)\times 1.60\mathrm{s}\\ =10.01 \mathrm{m}\end{array}$

Hence, the jumper lands just on the opposite bank of the river, neither long nor short.