Asked by vanny
the maximum height attained by the projectile motion is 600m and the range is 500m find
a. angle of projection
b. velocity of projection
c. total time of flight
a. angle of projection
b. velocity of projection
c. total time of flight
Answers
Answered by
Damon
Vertical problem:
at max height (half the trip)
600 = Vi t - .5*9.8 t^2
.5*9.8 t^2 - Vi t + 600 = 0
also
v = 0
so
0 = Vi -gt
t = Vi/g = Vi/9.8
so
.5*9.8(Vi^2/9.8^2) - Vi^2/9.8 +600 = 0
600 = .5*VI^2/9.8
Vi = 108 m/s initial up speed
t = rising time = 108/9.8 = 11.1 s
total time = 22.2 s
u = 500/22.2 = 22.5 m/s
tan theta = Vi/u = 108/22.5
= 4.8
theta = 78.2 degrees
speed = sqrt(108^2+22.5^2)
check my arithmetic !
at max height (half the trip)
600 = Vi t - .5*9.8 t^2
.5*9.8 t^2 - Vi t + 600 = 0
also
v = 0
so
0 = Vi -gt
t = Vi/g = Vi/9.8
so
.5*9.8(Vi^2/9.8^2) - Vi^2/9.8 +600 = 0
600 = .5*VI^2/9.8
Vi = 108 m/s initial up speed
t = rising time = 108/9.8 = 11.1 s
total time = 22.2 s
u = 500/22.2 = 22.5 m/s
tan theta = Vi/u = 108/22.5
= 4.8
theta = 78.2 degrees
speed = sqrt(108^2+22.5^2)
check my arithmetic !