Acceleration due to Gravity on Mars

[mlee16]

Homework Statement

Question: In 2004 two Martian probes successfully landed on the Red Planet. The final phase of the landing involved bouncing the probes until they came to rest (they were surrounded by protective inflated "balloons"). During one of the bounces, the telemetry (electronic data sent back to Earth) indicated that the probe took off at 25.0 m/s at an angle of 20 degrees and landed 110 m away (and then bounced again). Assuming the landing region was level, determine the acceleration due to gravity near the Martian surface.

Homework Equations

delta x = .5 * (velocity final + velocity initial) * time
Y final = y initial + .5 (velocity final + velocity initial) * time
Y final = y initial + (velocity initial) * time - .5 (acceleration) (time)^2

The Attempt at a Solution

delta x = .5 * (velocity final + velocity initial) * time
110 m = .5 * (25 cos 20 deg + 25 cos 20 deg) * time
Time = 4.68 s

I then found the height at the maximum point
Y final = y initial + .5 (velocity final + velocity initial) * time
Y final = 0 m + .5 (0 m/s + 25 sin 20 deg) * 4.68 sec
Y final = 20 m

I then used this equation to find acceleration
Y final = y initial + (velocity initial) * time - .5 (acceleration) (time)^2
20 = 0 m + (25 sin 20)(4.68) - .5(acceleration)(4.68)^2
Acceleration = 1.827 m/sec^2

 

[darkxponent]

Solution looks good. Doesn't it match with the answer?

 

[mlee16]

No haha I actually just figured it out. I should have halved my time when using the second equation because you're calculating the peak height.

 

[darkxponent]

The equatins you used are not generally used. But they are correct. I would suggest you remember the formulaes of projectile motion. You can derive it if you want. There are general formulaes for maximum height, range and time of flight!