Question

practice it
use the worked - example above to help you solve this problem. a long jumper (as shown in the figure) leaves the ground at an angle of 17.0° to the horizontal and at a speed of 10.8 m/s.
(a) how long does it take for her to reach maximum height?
0.322 s
(b) what is the maximum height?
0.56 m
the response you submitted has the wrong sign. m
(c) how far does she jump? (assume that her motion is equivalent to that of a particle, disregarding the motion of her arms and legs.)
6.65 m

Explanation:

Step1: Analyze vertical - initial velocity

The initial velocity $v_0 = 10.8\ m/s$ and the launch - angle $\theta=17.0^{\circ}$. The initial vertical velocity is $v_{0y}=v_0\sin\theta$. So, $v_{0y}=10.8\sin(17.0^{\circ})\ m/s$.

Step2: Find time to reach maximum height

At maximum height, the vertical velocity $v_y = 0$. Using the kinematic equation $v_y=v_{0y}-gt$, where $g = 9.8\ m/s^2$. Solving for $t$, we get $t=\frac{v_{0y}}{g}$. Substituting $v_{0y}=10.8\sin(17.0^{\circ})$ into the formula, $t=\frac{10.8\sin(17.0^{\circ})}{9.8}\ s\approx0.322\ s$.

Step3: Calculate maximum height

Using the kinematic equation $v_y^2 - v_{0y}^2=-2gh$. Since $v_y = 0$ at maximum height, $h=\frac{v_{0y}^2}{2g}$. First, $v_{0y}=10.8\sin(17.0^{\circ})\approx3.16\ m/s$. Then $h=\frac{(10.8\sin(17.0^{\circ}))^2}{2\times9.8}\ m=\frac{3.16^2}{19.6}\ m\approx0.51\ m$.

Step4: Find time of flight

The time of flight $T = 2t$ (because the motion is symmetric). Since $t=\frac{v_{0y}}{g}$, $T=\frac{2v_0\sin\theta}{g}$.

Step5: Calculate horizontal range

The horizontal velocity $v_{0x}=v_0\cos\theta$. The horizontal range $R = v_{0x}T$. Substituting $v_{0x}=v_0\cos\theta$ and $T=\frac{2v_0\sin\theta}{g}$, we get $R=\frac{v_0^2\sin2\theta}{g}$. Here, $\theta = 17.0^{\circ}$, $v_0 = 10.8\ m/s$, and $g = 9.8\ m/s^2$. So, $R=\frac{10.8^2\sin(2\times17.0^{\circ})}{9.8}\ m\approx6.65\ m$.

Answer:

(a) $0.322\ s$
(b) $0.51\ m$
(c) $6.65\ m$