Question

The froghopper, a tiny insect, is a remarkable jumper. suppose you raised a colony of the little critters on titan, a moon of saturn, where the acceleration due to gravity is only 1.35 m/s2, compared to the earth value of g = 9.81 m/s2. if on earth a froghopper\'s maximum height is h and maximum horizontal range is r, what would its maximum height and range be on titan in terms of h and r? assume the froghopper\'s takeoff velocity is the same on titan as on earth.

Answer 1

In projectile motion we know that range of projectile is given as

$R = \frac{v^2 sin2\theta}{g}$

also the maximum height is given as

$H = \frac{v^2sin^2\theta}{2g}$

so here we can see that maximum height and range on earth will inversely depends on gravity

now if we compare this height and range on earth and moon

$\frac{R_E}{R_M} = \frac{g_m}{g_e}$

$\frac{R}{R_M} = \frac{1.35}{9.8}$

$R_M = \frac{R* 9.8}{1.35} = 7.26 R$

similarly for height we have

$\frac{H_E}{H_M} = \frac{g_m}{g_e}$

$\frac{H}{H_M} = \frac{1.35}{9.8}$

$H_M = \frac{H* 9.8}{1.35} = 7.26 H$

so both will increase by factor of 7.26

Answer 2

The height to which the frog-hopper jumps on the surface of titan will be $\boxed{7.26h}$ and the range of the frog-hopper will be $\boxed{7.26R}$.

Explanation:

The motion of the projectile on the surface of the planet occurs in two dimension. The vertical motion of the frog-hopper will be under the action of the gravity whereas the horizontal motion of the frog-hopper will be the motion under constant speed.

Write the expression for the maximum height attained by the frog-hopper:

$h=\dfrac{v_{y}^2}{2g}$                                               ...... (1)

Here, $h$ is the maximum height attained by the frog-hopper, $v_y$ is the velocity in vertical direction.

Consider that the height attained by the frog-hopper on the surface of titan is $h_{T}$ and $h_{E}$ on the surface of Earth.

Compare the maximum heights attained by the frog-hopper on Earth and titan:

$\dfrac{h_{E}}{h_{T}}=\dfrac{g_T}{g_E}$

Substitute the value of the acceleration due to gravity on each surfaces.

$\begin{aligned}\dfrac{h}{h_T}&=\dfrac{1.35\text{ m/s}^2}{9.81\text{ m/s}^2}\\h_T&=7.26h\end{aligned}$

Write the expression for the range of jump of the frog-hopper:

$R=\dfrac{v^2sin2\theta}{g}$                                                    ...... (2)

Compare the range of the frog-hopper on the surface of Earth and titan:

$\dfrac{R_{E}}{R_{T}}=\dfrac{g_T}{g_E}$

Substitute the value of the acceleration due to gravity on each surfaces.

$\begin{aligned}\dfrac{R}{R_T}&=\dfrac{1.35\text{ m/s}^2}{9.81\text{ m/s}^2}\\R_T&=7.26R\end{aligned}$

Thus, the height to which the frog-hopper jumps on the surface of titan will be $\boxed{7.26h}$ and the range of the frog-hopper will be $\boxed{7.26R}$.

Learn More:

1. The acceleration of the block on friction surface brainly.com/question/7031524

2. The one which is not a component of a lever brainly.com/question/1073452

3. Change in the momentum of the car due to the collision brainly.com/question/9484203

Answer Details:

Grade: High School

Subject: Physics

Chapter: Projectile motion

Keywords:

froghopper, tiny insect, titan, range, maximum height, vertical direction, projectile, in terms of h and r, takeoff velocity, raised a colony.