At a certain billboard, a coin was thrown horizontally and reached 16 meters, from the base of the billboard 3.5 seconds after it was thrown. Calculate the following:
A. Velocity when it was thrown
B. Velocity after 2.5 seconds
C. Height it was thrown.
Horizontal distance, H = 16 m
Time of flight, T = 3.5 s
Horizontal velocity, Vh = 16/3.5 = 4.57 nm/s
Time of flight, T = 3.5 s
Therefore Coin reaches maximum height, h at t = T/2 = 1.75 s
Vertical velocity v will be zero at max height.
Using v = u + at,
0 = Vu – 9.8*1.75 (where Vu is the initial vertical velocity)
Vu = 17.15 m/s
Velocity of Projection
V^2 = Vh^2 + Vu^2
V^2 = 4.57^2 + 17.15^2
V^2 = 315
V = 17.75 m/s
Vertical velocity
Using v = u + at
After 2.5 s of flight
Vv = Vu – gt
Vv = 17.15 – 9.8*2.5
Vv = -7.35 m/s (negative velocity = downwards direction)
Resultant velocity
V_(2.5)^2 = Vh^2 + Vv^2
V_(2.5)^2 = 4.57^2 + (-7.35)^2
V_(2.5)^2 = 74.9
V_(2.5) = 8.65 m/s
Max Height
Using s = ut + (1/2)at^2
h = Vu*(T/2) – (g/2)(T/2)^2 (max height reached at t = T/2)
h = 17.15*1.75 – 4.9*(1.75)^2
h = 30.0125 – 15.00625
h = 15 m