Question #304549

The acceleration due to gravity at the surface of Mars is roughly 2/5 of Earth’s “g”. If an astronaut on Mars were to toss a wrench upward with a speed of 10 m/s, find a) how long it would rise, b) how high it would go, c) its speed at t = 3.0s , and d) its displacement at t = 3.0s.


Expert's answer

Using the kinematic formula (see https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas), obtain the following.

a) Time of rise as the time required to come to zero velocity at the highest point:


t=v0gt = \dfrac{v_0}{g}

where v0=10m/s,g=25⋅9.8m/s2=3.92m/s2v_0 = 10m/s,g = \dfrac25\cdot9.8m/s^2 =3.92m/s^2. Thus, obtain:


t=1023.92≈26st = \dfrac{10^2}{3.92} \approx 26s

b) Distance, travelled by this time, which is the maximum height:


h=v0t2=10⋅262=130mh = \dfrac{v_0t}{2} = \dfrac{10\cdot 26}{2} = 130m

c) Speed at t=3s:


v=∣v0−gt∣=∣10m/s−3.92m/s2⋅3s∣≈1.8m/sv = |v_0-gt|= |10m/s - 3.92m/s^2\cdot 3s|\approx 1.8m/s

d) Displacement at t=3s:


d=v0t−at22=10m/s⋅3s−3.92m/s2⋅(3s)22≈12md = v_0t-\dfrac{at^2}{2} = 10m/s\cdot 3s - \dfrac{3.92m/s^2\cdot (3s)^2}{2}\approx 12m
LATEST TUTORIALS
APPROVED BY CLIENTS