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.
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
