a projectile is shot from the edge of a clif h=205m above the ground level with
ID: 1696356 • Letter: A
Question
a projectile is shot from the edge of a clif h=205m above the ground level with an initial speed of vo=125 m/s at an angle of 37.0 degrees with the horizontal.
a) determine the time taken by the projectile to hit point P at ground level
b) determine the range X of the projectile as measured from the base of the cliff (km)
c) at the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity
horizontal-
vertical-
d) what was the magnitude of the velocity?
e) what is the angle made by the velocity vector with the horizontal? (below the horizontal)
f) find the maximum height above the cliff top reached by the projectile
Explanation / Answer
Given
Height of the cliff = 205 m
initial speed = v0 = 125 m/s
angle = 37
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(a)
Time of descent = sqrt(2h/g)
= sqrt(2*205/9.8)
= 6.46 sec
(b)
Range from the base of the cliff = R = u[sqrt(2h/g)]
=125*6.46
= 808.5 m
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(c)
horizontal component x = ux = ucos
= 125*cos(37)
= 99.8 m/sec
vertical component of y = uy = usin
= 125*sin(37)
= 75.22 m/sec
_______________________________________________________________________________
(d)
magnitude of velocity v = sqrt(vx^2+vy^2)
= sqrt[(99.8)^2+(75.22)^2)]
= 124.97 m/s
__________________________________________________________________________________
(e)
tan = vy/vx
= tan^-1(75.22/99.82)
= 37
_____________________________________________________________________________________
(f)
maximum height = u^2sin^2/2g
= (125)^2(sin(37)^2)/2*9.8
= 288.72 m
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