Martian Projectile Motion: Finding Maximum Height and Range

[Atreyu]

Homework Statement

A Martian creature similar to an Earth frog jumps with an initial
speed v0 and attains the range R over horizontal ground.
The maximum possible height reached by the creature, neglecting
friction in the tenuous Martian air, is ($$\vartheta$$ being the
launch angle):

(a) R/4tan$$\vartheta$$;
(b) R/4sin$$\vartheta$$;
(c) R/2tan$$\vartheta$$;
(d) undetermined because of missing data.

Homework Equations

I tried using some formulas to derive the answer...

$$\frac{voy}{g}$$=t_{highest point}

and solved for t,

t=$$\frac{voysin\vartheta}{g}$$

t=$$\frac{R}{vcos\vartheta}$$

and tried using a variety of the range projectile motion equation to get to this answer

The Attempt at a Solution

I know the answer is (a), but is there any easy way to geth there, I feel like I've been going around in circles

 

[mark726]

.





Thank you for your question. It seems like you are on the right track with your attempt at a solution. To find the maximum height reached by the Martian creature, we can use the equation for the vertical component of the creature's motion, which is given by:

y(t) = y0 + v0y*t - 0.5*g*t^2

Where y0 is the initial height (which we can assume to be 0 since the creature is jumping from the ground), v0y is the initial vertical velocity (which is given by v0*sin\vartheta), and g is the acceleration due to gravity on Mars (which we can assume to be constant at 3.711 m/s^2).

We want to find the maximum height, so we need to find the time t at which the creature reaches its highest point. This can be done by setting the derivative of y(t) with respect to t equal to 0 and solving for t. This gives us:

t = v0*sin\vartheta/g

Now, we can substitute this value of t into the equation for y(t) to find the maximum height reached by the creature:

ymax = v0^2*sin^2\vartheta/2g

We also know that the range R is given by:

R = v0^2*sin2\vartheta/g

Rearranging this equation, we get:

v0^2 = Rg/sin2\vartheta

Substituting this value of v0^2 into the equation for ymax, we get:

ymax = (Rg/sin2\vartheta)*sin^2\vartheta/2g

Simplifying this, we get:

ymax = R/4tan\vartheta

Therefore, the correct answer is (a). I hope this helps you understand how to approach this problem. Keep up the good work!