91Ó°ÊÓ

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 $\mathrm{θ}=45°$.

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.

Part (a)

The range of a projectile motion is

$R=\frac{v_o^2\sin 2\mathrm{θ}}{g}$.

Now, for the first part of the motion,

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

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

Part (b)

For the vertical motion,

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

Now, calculating the value of t,

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

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

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

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