a stone is thrown vertically upward with a speed of 10.5 m/s from the edge of a cliff 76.0 m high.
how much later does it reach the bottom of cliff?
what is its speed just before hitting?
what total distance did it travel
V = Vo + g*Tr = 0
10.5 - 9.8*Tr = 0
9.8t = 10.5
Tr = 1.07 s. = Rise time.
hmax = ho + (V^2-Vo^2)/2g =
76 + (0-10.5^2)/-19.6 = 81.63 m. Above gnd.
hmax = Vo*t + 0.5g*t^2 = 81.63 m.
0 + 9.8*t^2 = 81.63
t^2 = 8.33
Tf = 2.89 s. = Fall time.
Tr+Tf = 1.07 + 2.89 = 3.96 s. To reach
bottom of cliff.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*81.63 = 1599.95
V = 40 m/s Just before hitting gnd.
d = (hmax-ho)+ hmax
d = (81.63-76) + 81.63 = 87.26 m. =
Total distance traveled.
To answer the given questions, we can use the equations of motion in vertical motion. Let's break down the problem and apply the equations step by step.
1. How much later does it reach the bottom of the cliff?
First, we need to find the time it takes for the stone to reach its peak height. The initial velocity when thrown vertically upward is 10.5 m/s, and the height of the cliff is 76.0 m.
Using the equation for vertical displacement (y) in free fall:
y = voy * t + (1/2) * a * t^2
Since the stone is thrown upward, the initial velocity (voy) is positive, and the acceleration due to gravity (a) is -9.8 m/s² (negative because it acts downward).
When the stone reaches its peak, its final velocity (vfy) will be 0 m/s. Therefore:
0 = 10.5 m/s - 9.8 m/s² * t1
Solving for t1, we find:
t1 = 10.5 m/s / 9.8 m/s² ≈ 1.07 seconds
Next, to find the time it takes to reach the bottom of the cliff, we can double the time it took to reach the peak because the total time of ascent equals the total time of descent.
Therefore, the time it takes for the stone to reach the bottom of the cliff is:
t_total = 2 * t1 ≈ 2.14 seconds
So, the stone reaches the bottom of the cliff approximately 2.14 seconds later.
2. What is its speed just before hitting?
Now that we know the time it takes for the stone to reach the bottom of the cliff, we can find its final velocity.
Using the equation for vertical velocity:
vf = voy + a * t_total
Substituting the values:
vf = 10.5 m/s - 9.8 m/s² * 2.14 s
Calculating vf, we find:
vf ≈ -9.03 m/s
The negative sign indicates that the stone is moving downward.
Therefore, the speed of the stone just before hitting the bottom of the cliff is approximately 9.03 m/s.
3. What total distance did it travel?
To find the total distance traveled by the stone, we can combine the distances traveled during ascent and descent.
During ascent:
Distance_asc = voy * t1 + (1/2) * a * t1^2
During descent:
Distance_desc = vf * t_total + (1/2) * a * t_total^2
Since the stone moves upward and then downwards, we must account for both distances. However, we need to consider the magnitudes of these distances as distance is a scalar quantity.
Therefore, the total distance traveled is:
Total distance = |Distance_asc| + |Distance_desc|
Substituting the values and calculating, we find:
Total distance = |10.5 m/s * 1.07 s + (1/2) * (-9.8 m/s²) * (1.07 s)^2| + |-9.03 m/s * 2.14 s + (1/2) * (-9.8 m/s²) * (2.14 s)^2|
Calculating the total distance, we find:
Total distance ≈ 76 m + 82 m
Therefore, the total distance traveled by the stone is approximately 158 meters.