Question

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has 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. We can ignore air resistance.

Answer 1

Answer:

t = 2.32 s

Explanation:

Applying the equation of motion;

d = ut + 0.5gt^2

Where;

d = distance travelled

u = initial velocity

g = acceleration due to gravity

t = time taken

Since the object was dropped;

u = 0

Then,

d = 0.5gt^2

t^2 = d/0.5g

t = √(d/0.5g) .......1

Given

g = 3.7 m/s^2

d = 10 m

Substituting the values;

t = √(10/(0.5×3.7))

t = 2.32 s

Answer 2

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of  the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.