Answer:
3.68 m/s
Corrected question;
Chef Andy tosses an orange in the air, then catches it again at the same height. The orange is in the air for 0.75 s. We can ignore air resistance. What was the orange's velocity at the moment it was tossed into the air?
Explanation:
Applying the equation of motion;
v = u + at. ....1
Where
u = initial speed
v = final speed
a = acceleration
t = time taken
Given;
Total time of flight t = 0.75s
Acceleration due to gravity g = 9.8 m/s^2
Considering the first phase of flight, during the upward motion to the maximum height. The final velocity at the maximum height is zero and the time of flight of the phase is equal to half of the total flight time.
v1 = 0
t1 = t/2 = 0.75/2 =0.375s
From equation 1;
v1 = u - gt1
Substituting the values;
0 = u - 9.8(0.375)
Solving for u;
u = 9.8(0.375)
u = 3.675
u = 3.68 m/s
Initial velocity = 3.68 m/s
