a projectile is thrown from the ground with an initial velocity of 20.0m/s at an angle of 40 degrees above the horizontal, find the maximum height,the time required to reach its maximum height,c) it velocity at the top of the trajectory.d) the range of the projectile,e) the total time of the flight

Answer 1

Vo = 20m/s[40o]
Xo = 20*Cos40 = 15.32 m/s.
Yo = 20*sin40 = 12.86 m/s.

a. Y^2 = Yo^2 + 2g*h
Y = 0
Yo = 12.86 m/s.
g = -9.8 m/s^2
Solve for h.

b. Y = Yo + g*Tr
Y = 0
Yo = 12.86 m/s.
g = -9.8 m/s^2.
Solve for Tr.(Rise time or time to reach max. ht.)

c. V = Xo + Yi = Xo + 0 = Xo

d. Range = Vo^2*sin(2A)/g
Vo = 20 m/s
A = 40o
g = 9.8 m/s^2.
Solve for Range.

e. Tf = Tr = Fall time.
T = Tr+Tf = 2Tr = Time in flight.