A bullet fired at an angle θ = 60o with a velocity of 20 m/s. Acceleration due to gravity is 10 m/s2. What is the time interval to reach the maximum height?
Answers
Known :
The initial velocity of bullet (vo) = 20 m/s
Angle (θ) = 60oC
Acceleration due to gravity (g) = 10 m s–2
Wanted : The time interval to reach the maximum height
Solution :
The initial velocity at the horizontal direction (x axis) :
vox = vo cos 60o = (20)(0.5) = 10 m/s
The initial velocity at the vertical direction (y axis) :
voy = vo sin 60o = (20)(0.5√3) = 10√3 m/s
The time interval to reach the maximum height, calculated using this equation :
vty = voy + g t
vty = the final velocity in the vertical direction = the final velocity at the highest point = 0 m/s
voy = the initial velocity at the horizontal direction = 10√3 m/s
g = acceleration due to gravity = 10 m/s2
t = time interval
The time interval :
vty = voy + g t
0 = 10√3 – 10 t
10√3 = 10 t
t = 10√3 / 10
t = √3 seconds