A rock is thrown from a cliff top at 18m/s, 25 degrees above the horizontal. It lands on the beach 4.2 seconds later.
a) What is the height of the cliff?
b) Hor far from the base of the cliff did the rock land?
vertical veloctiy: 18sin25
Hf=hi+viv*t-1/2 g t^2
hf=0, hi=unknown,viv=18sin25, t=4.2 solve for hi.
horizontal distance? 18cos25*4.2
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To find the height of the cliff (a), we can use the equations of motion and break down the initial velocity of the rock into its vertical and horizontal components.
a) The vertical component of the initial velocity can be found using the equation vₒy = vₒsin(θ), where vₒ is the initial velocity and θ is the angle above the horizontal.
vₒy = 18m/s * sin(25°)
≈ 7.58 m/s
Next, we can use the vertical equation of motion, h = vₒy * t + 0.5 * g * t², where h is the height of the cliff and g is the acceleration due to gravity (approximately -9.8 m/s²).
Substituting the values, we get:
h = (7.58 m/s)(4.2 s) + 0.5 * (-9.8 m/s²)(4.2 s)²
≈ 31.90 m
Therefore, the height of the cliff is approximately 31.90 meters.
b) To find the horizontal distance the rock landed from the base of the cliff, we can use the horizontal component of the initial velocity, which can be found using the equation vₒx = vₒcos(θ), where vₒ is the initial velocity and θ is the angle above the horizontal.
vₒx = 18m/s * cos(25°)
≈ 16.17 m/s
The horizontal distance traveled can be calculated using the equation d = vₒx * t, where d is the horizontal distance and t is the time taken for the rock to land.
d = (16.17 m/s)(4.2 s)
≈ 67.914 m
Therefore, the rock landed approximately 67.914 meters from the base of the cliff.