Question

A projectile is fired with an initial velocity of 75.0 m/s at an angle of 33.0 degrees above the horizontal. It lands on a hillside 50m above the level from which it was fired. Determine 1) the time of flight of the projectile. 2) the maximum height to which it rises. 3) the horizontal distance from launching to landing points.

Answer #1

given

initial velocity u = 75 m/s

theta = 33 degrees

final verticla height y = 50 m

a) from the equation

y-y0 = usin(theta)*t-1/2gt^2

y0 = 0

50 = 75sin(33)*t-1/2*9.8t^2

4.9t^2-40.85t+50 = 0

this is quadratic equation

it looks like ax^2+bx+c = 0

x = -b+-sqrt(b^2-4ac)/2a

time has two roots

t = 6.85 sec and 1.49 sec

2) h max = u^2sin^2(theta)/2g

h max = 75^2sin^2(33)/2*9.8

h max = 85.13 m

c) horizontal disatnce

x = ucos(theta)*t

for t = 6.85 sec

x = 75cos(33)*6.85 = 430.8 m

for t = 1.49 sec

x = 75cos(33)*1.49 = 93.72 m

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