Question

One drops a basketball from a height of 10m on mars, where the acceleration due to gravity has a magnitude of 3.7 m/s^2.

We want to know how many seconds the basketball is in the air before it hits the ground.
a) 1.0 s
b) 1.5 s
c) 2.0 s
d) 2.5 s

Which kinematic formula would be most useful to solve for the target unknown?
a) s=(v_0)t + (1/2)at^2
b) v=(v_0) + at
c) v^2=(v_0)^2 + 2as
d) s=(v_0)t − (1/2)at^2

Answer 1

The basketball is in the air for 2.5 seconds before it hits the ground on Mars.

Step-by-step explanation:

To calculate the time a basketball dropped from 10m on Mars is in the air, we use the kinematic equation s = (v_0)t + (1/2)at². By substituting the known values into the equation, solving for t will provide the duration before the basketball hits the ground.

To calculate the time it takes for the basketball to hit the ground on Mars, we can use the kinematic formula: s = (v0)t + (1/2)at2, where s is the initial height, v0 is the initial velocity (which is 0 in this case), a is the acceleration due to gravity on Mars (-3.7 m/s2), and t is the time. Plugging in the values, we get 10 = (0)t + (1/2)(-3.7)t2. By solving this quadratic equation, we find that t = 2.5 s. Therefore, the correct answer is (d) 2.5 s.