A projectile is fired at an angle of 45 degrees to the horizontal. If its initial velocity is 90 m/s, how far does the projectile travel? (Assume g = 10m/s^2.)
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Expert Answers
calendarEducator since 2010
write8 answers
starTop subject is Science
First you need to find the time of flight:
y = v_yo t + (1/2) g t^2
0 = (90 sin 45) t + (5) t^2
factor out a t
0 = t (90 sin45 + 5 t)
set roots equal to zero
t = 0 and 63.6 + 5 t = 0
t = 63.6/5 = 12.7 sec
To find distance use
x = v_xo t
x = 90 cos 45 * 12.7
x = 808 m
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calendarEducator since 2010
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The projectile is fired with a velocity of magnitude 90m/s at angle 45 degree to the horizontal. We can divide the initial velocity into a horizontal component of 90 sin 45 = 90/sqrt 2 and a vertical component of 90 cos 45 = 90/sqrt 2.
The downward acceleration due to gravity of 10 m/s^2 acts on the vertical component and reduces it. The projectile reaches a vertical velocity equal to 0 at the highest point and when it returns to ground level, the magnitude of velocity is 90/sqrt 2 though it is in the opposite direction. We can use this to calculate the time that the projectile was in motion.
- 90/sqrt 2 = 90/sqrt 2 – 10*t
=> 2*90/ sqrt 2 = 10*t
=> t = 2*9/sqrt 2
=> t = 18/sqrt 2
The horizontal distance travelled with a velocity 90/sqrt 2 in 18/sqrt 2 sec is equal to
(90/sqrt 2)*(18/sqrt 2) = 90*18 / 2 = 90*9 = 810 m.
Therefore the projectile travels a horizontal distance of 810 m.