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\,\text s0.75s0, point, 75, start text, s, end text. We can ignore air resistance. What was the orange's velocity at the moment it was tossed into the air?

3.68 m/s

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

f=ma newton's second law 7x8=56n