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
(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
exercise
a grasshopper jumps a horizontal distance of 1.50 m from rest, with an initial velocity at a 47.0° angle with respect to the horizontal.
(a) find the initial speed of the grasshopper.
enter a number m/s
(b) find the maximum height reached.
m

Explanation:

Step1: Analyze vertical - motion for grasshopper (a)

The horizontal range formula is $R=\frac{v_0^{2}\sin2\theta}{g}$. We need to solve for $v_0$. Given $R = 1.50$ m and $\theta=47.0^{\circ}$, and $g = 9.8$ m/s². Rearranging the formula for $v_0$ gives $v_0=\sqrt{\frac{Rg}{\sin2\theta}}$.

Step2: Calculate $v_0$

First, find $\sin2\theta=\sin(2\times47.0^{\circ})=\sin94^{\circ}\approx0.9976$. Then $v_0=\sqrt{\frac{1.50\times9.8}{0.9976}}=\sqrt{\frac{14.7}{0.9976}}\approx\sqrt{14.74}\approx3.84$ m/s.

Step3: Analyze vertical - motion for grasshopper (b)

The formula for the maximum height in projectile motion is $H=\frac{v_0^{2}\sin^{2}\theta}{2g}$. We know $v_0\approx3.84$ m/s and $\theta = 47.0^{\circ}$, so $\sin\theta=\sin47.0^{\circ}\approx0.7314$.

Step4: Calculate $H$

$H=\frac{(3.84)^{2}\times(0.7314)^{2}}{2\times9.8}=\frac{14.7456\times0.535}{19.6}=\frac{7.88}{19.6}\approx0.40$ m.

Answer:

(a) 3.84 m/s
(b) 0.40 m