A projectile is fired with initial velocity 95 m/s at an angle of 50 degrees from the horizontal. After 5 s, it strikes the top of a hill. What is the elevation of the hill above the point of firing?
Answers
Explanation:
Calculating the time taken by the projectile to reach the ground:
s=ut+
2
1
gt
2
500m=(100m/s)t+
2
1
(9.8m/s
2
)t
2
500=100t+4.9t
2
t=4.15s
Calculating the distance of the target from the hil:
Range = ut=(100m/s)(4.15s)=415m
Calculating the velocity with which the projectile hits the ground:
v=100m/s+(9.8m/s
2
)(4.15s)
2
=268.78m/s
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21.27 degrees
Given:
A projectile is fired with initial velocity 95 m/s at an angle of 50 degrees from the horizontal
To find:
elevation of the hill above the point of firing
Solution:
Vertical acceleration a = -9.8 m/s^2.
Launch speed S = 95 m/s at 50° elevation.
As function of time, vertical velocity v = u + a*t
where, initial value u = Ssin50 = 72.77 m/s.
To reach apex where v = 0,
elapsed time, t = -u/a = -72.77/-9.8 = 7.43 s.
But for 5 s flight, projectile never reaches apex, so impact time T = 5 s.
At impact, height =
= 72.77*5 -9.8/2*25 = 363.9 -122.5
= 241.4 m.
Horizontal range = S*cos50*T = 305.3 m.
Line of sight displacement
= = 389.2 m.
At impact, vertical velocity v = u + a*T
= 72.77 -9.8*5 = 23.77 m/s
Horizontal velocity w = S*cos50 = 61.06 m/s
Impact speed =
=
= 65.53 m/s.
Impact elevation angle = arctan(v/w)
= arctan(23.77/61.06)
= 21.27°
At impact after 5 seconds:
height 241.4m, range 305.3m, displacement 389.2m, velocity 65.53 m/s at 21.27 degrees, still rising.
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