A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
e base of the cliff.
1 answer:
Answer:
The rock must leave the cliff at a velocity of 28.2 m/s
Explanation:
The position vector of the rock at a time t can be calculated using the following equation:
r = (x0 + v0x · t, y0 + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.
When the rock reaches the ground, the position vector will be (see r1 in the figure):
r1 = (90 m, -50 m)
Then, using the equation of the vector position written above:
90 m = x0 + v0x · t
-50 m = y0 + 1/2 · g · t²
Since x0 and y0 = 0:
90 m = v0x · t
-50 m = 1/2 · g · t²
Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:
-50 m = 1/2 · g · t²
-50 m = -1/2 · 9.81 m/s² · t²
-50 m / -1/2 · 9.81 m/s² = t²
t = 3.19 s
Now, using the equation of the x-component of r1:
90 m = v0x · t
90 m = v0x · 3.19 s
v0x = 90 m / 3.19 s
v0x = 28.2 m/s
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Answer:
1) The time it takes the diver to move off the end of the diving board to the pool surface, , is approximately 1.392 seconds
2) The horizontal distance from the edge of the pool to where the diver enters the water, , is approximately 5.76 meters
Explanation:
1) The given parameters are;
The height of the diving board above the pool's surface, h = 9.5 m
The length by which the diving board over hangs the pool L = 2 m
The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s
Taking = The time it takes the diver to move off the end of the diving board to the pool surface
Therefore, we have from the equation of free fall;
h = 1/2 × g × ²
Where;
g = The acceleration due to gravity = 9.81 m/s²
Substituting the values, gives;
9.5 = 1/2 × 9.81 × ²
= √(9.5/(1/2 × 9.81)) ≈ 1.392 s
The time it takes the diver to move off the end of the diving board to the pool surface = ≈ 1.392 s
2) The horizontal distance, , in meters from the edge of the pool to where the diver enters the water is given as follows;
= L + v₀ ×
= 2 + 2.7× 1.392 ≈ 5.76 m
∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.