Question

a sandwich is thrown upwards by a street vendor with velocity 10 m/s to a person in the second floor window of a building. if the vendor releases the sandwich from 2 m above the ground and the second floor window is 5 m above the ground at what velocity will the sandwich be moving when it is caught?

Answer 1

The distance flown by sandwich is

H = 5[m] - 2[m] = 3[m].

Let the time needed the sandwich to fly this distance be T. Then

H = VT - gT²/2,

or

3[m] = 10[m/s]*T - 9.8[m/s²]*T²/2.

Let's find T.

4.9T² - 10T + 3 = 0 ==> T ≈ min{ 1.68, 0.37 } = 0.37 [s].

Therefore, the velocity of the sandwich at the moment of catching is

V ≈ 10[m/s] - 9.8[m/s²]*0.37[s] = 6.374 [m/s].