vertical problem:
goes up 150 m (remarkable throw)
v = Vi - g t
v = 0 at top
Vi = g t = initial up component of speed
h = Vi t - (1/2) g t^2
150 = Vi t - (1/2) g t^2
150 = g t^2 -(1/2) g t^2
300 = g t^2
300/9.81 = t^2
t = 5.53 seconds upward, 11.1 total in air
Vi = 9.81*5.53 = 54.2 meters/sec Upward
= 100 sin angle up from horizon
so angle = sin^-1(54.2/100)
= 32.8 degrees up from level
for range, horizontal problem
speed horizontal = u = 100 cos32.8 forever
u = 84.1 m/s
time in air = 11.1 seconds
so
range = 84.1*11.1 meters
(I think it was an arrow if not a bullet)
A ball thrown with a speed of 100m/s attains a height of 150.calculate
(a)the time of flight
(b)the angle of projection
(c)the range
2 answers
Time of flight:
That would be twice the time it takes the ball to fall H = 150 m, which is
t = 2*sqrt(2H/g) = 11.07 s
(You double the fall time to get the total time of flight)
Launch angle:
V*sinA/g = t/2 = 5.533 s
sin A = 0.5422
A = 32.8 degrees
Range:
(Vo^2/g)*sin(2A) = 930 m
That would be twice the time it takes the ball to fall H = 150 m, which is
t = 2*sqrt(2H/g) = 11.07 s
(You double the fall time to get the total time of flight)
Launch angle:
V*sinA/g = t/2 = 5.533 s
sin A = 0.5422
A = 32.8 degrees
Range:
(Vo^2/g)*sin(2A) = 930 m