practice it\nuse the worked - example above to help you solve this problem. a long jumper (as shown in the…

practice it\nuse 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.\n(a) how long does it take for her to reach maximum height?\n0.322 s\n(b) what is the maximum height?\n0.56 m\nthe response you submitted has the wrong sign. m\n(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.)\n6.65 m

practice it\nuse 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.\n(a) how long does it take for her to reach maximum height?\n0.322 s\n(b) what is the maximum height?\n0.56 m\nthe response you submitted has the wrong sign. m\n(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.)\n6.65 m

Answer

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) $6.65\ m$