A boy wishes to throw a rock into the river below the cliff where he is standing. The cliff is 50 M high and he can throw with a horizontal velocity of 30 m/s. how long does it take for the rock to reach the water? How far, in the horizontal direction, does the rock land?
You are looking for range. Range = Vh x t
Vh = 30 m/s
t= ?
We know d= 50 m, viv = 0 m/s a=9.8
d=vit + 1/2 at ^2
50=0 + 1/2 9.8 t ^2
50 = 4.9 t^2
t = 3.1 s
Range = (30) (3.1)
= 95.8 m
Hope this helped !
To determine the time it takes for the rock to reach the water and the horizontal distance it lands, we can use the equations of motion.
First, let's find the time taken for the rock to fall from the top of the cliff to the water. We can use the vertical motion equation:
y = uyt + (1/2)gt^2
Where:
y = vertical displacement (height of the cliff) = 50 m
uy = initial vertical velocity = 0 (as only the horizontal velocity is given)
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time taken
Rearranging the equation, we have:
50 = 0*t + (1/2)*9.8*t^2
50 = 4.9t^2
Simplifying, we get:
t^2 = 50/4.9
t^2 ≈ 10.2
t ≈ √10.2
t ≈ 3.19 seconds
So, it takes approximately 3.19 seconds for the rock to reach the water.
Now, to find the horizontal distance traveled by the rock, we can use the horizontal motion equation:
x = uxt
Where:
x = horizontal displacement (distance traveled by the rock)
ux = initial horizontal velocity = 30 m/s (given)
t = time taken = 3.19 seconds (as calculated earlier)
Plugging in the values, we have:
x = 30 * 3.19
x ≈ 95.7 meters
Therefore, the rock lands approximately 95.7 meters horizontally from the cliff.