A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17.0 m/s

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17.0 m/s. The cliff is h = 61.0 m above a flat horizontal beach.



How long after being released does the stone strike the beach below the cliff?
_____s
With what speed and angle of impact does the stone land?
_____m/s
_____° below the horizontal

Answer 1

1. Vertical velocity [Vf^2-Vi^2=-2gX]

Vf^2-0=-2(9.8m/s^2)(-61m) **Vi=0 because ball thrown horizontally*
Vf=34.577m/s

2.Average vertical velocity [Vi+Vf]/2
Vavg= [0+34.577]/2
Vavg=17.2887m/s

3. Time taken to reach ground [t=X/Vavg]
t=61m/17.2887m/s= {3.529sec}

4. Net Vector hitting the ground [Vfx^2+Vfy^2]^(1/2)

Net Vector = (17^2+34.577^2)^(1/2)= {38.53m/s}

5.Angle hit ground [Tan^(-1) Vfy/Vfx]

Angle= tan^(-1) 34.77/17= {63.82 degrees}

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how long takes: {3.529sec}
With what speed and angle of impact does the stone land? 38.53m/s @ 63.82 degrees bellow the horizontal

Answer 2

You solve this by using treating the horizontal speed and vertical speed separately.

1. the horizontal speed is constant. If the stone is moving horizontally at 17 m/s at the start, it is still moving at this speed at the end.

2. the vertical speed is governed by the normal rules for accelerating towards the earth. It's exactly the same as if the stone was let fall vertically.

Answer 3

noting that the preliminary horizontal speed ingredient is 0, s is 50.7 and a = 9.8 the vertical speed ingredient is a crimson herring at this factor use s = ut + (a million/2) at^2

Answer 4

how long to reach the ground
s=d/t
a=(v-u)/t
t=d/s
s=(v+u)/2
a=(v-u)(v+u)/2d
v=0
a=-u^2/2d
a=9.81
d=61
9.81=u^2/122
1196.82=u^2
u=34.6m/s
average speed=(0+34.6)/2=17.3m/s
s=d/t
17.3=61/t
t=61/17.3=3.526s

time taken to hit the floor=3.526s

Answer 5

How much is wind load value?