For this problem, a person standing on the edge of a cliff throws a rock directly upward. it is observed that 2 secs later the rock is at its maximum height (in feet) and that 5 secs after that (meaning 7 seconds after being thrown) the rock hits the ground at the base of the cliff.
(a) What is the initial velocity of the rock? ft/sec.
(b) How high is the cliff? t ft.
(c) What is the velocity of the rock at time t? ft/sec.
(d)What is the velocity of the rock when it hits the ground? ft/sec.
a. V = Vo + g*t = 0
Vo - 32*2 = 0
Vo = 64 Ft/s.
b. Tf1 = Tr = 2 s. = Fall time from hmax
to top of cliff.
Tf2 = 7-Tr-Tf1 = 7 - 2 - 2 = 3 s. = Fall
time from top of cliff to gnd.
h = Vo*Tf2 + 0.5g*Tf2^2
h = 64*3 + 16*3^2 = 336 Ft. = Ht. of cliff.
c. What is time t?
d. h max = ho + -(Vo^2)/2g
h max = 336 + -(64^2)/-64 = 400 Ft. Above gnd.
V^2 = Vo^2 + 2g*h = 0 + 64*400 = 25,600
V = 160 Ft/s.
(a) The initial velocity of the rock can be determined using the fact that the time it takes for the rock to reach its maximum height is half the total time it takes for the rock to hit the ground.
So, the time it takes for the rock to reach its maximum height is 2 seconds, and the total time it takes for the rock to hit the ground is 7 seconds. Therefore, the initial velocity can be calculated as follows:
Initial velocity = (distance traveled during ascent) / (time taken for ascent)
= (0 ft - h ft) / 2 sec
= -h/2 ft/sec
where h is the maximum height reached by the rock.
(b) To determine the height of the cliff, we need to find the maximum height reached by the rock. We can use the following kinematic equation:
h = Initial velocity * time + 0.5 * acceleration * time^2
At maximum height, the velocity is zero. So, we can substitute the values:
0 = -h/2 ft/sec * 2 sec + 0.5 * (-32 ft/sec^2) * (2 sec)^2
0 = -h + 64
h = 64 ft
Therefore, the height of the cliff is 64 ft.
(c) The velocity of the rock at time t can be determined using the following equation:
Velocity = Initial velocity + acceleration * time
Here, at time t (which is after 2 seconds), the acceleration is -32 ft/sec^2 and the initial velocity is -h/2 ft/sec. So, we can substitute the values:
Velocity = -h/2 ft/sec + (-32 ft/sec^2) * (t - 2 sec)
(d) The velocity of the rock when it hits the ground is equal to its initial velocity plus the product of acceleration and the time taken for descent (t - 2 sec). So, we can substitute the values:
Velocity = -h/2 ft/sec + (-32 ft/sec^2) * (t - 2 sec)
Let's solve the problem step by step:
(a) To find the initial velocity of the rock, we can use the fact that the time it takes for the rock to reach its maximum height is 2 seconds. Since the rock is thrown directly upward, it will have zero velocity at its maximum height. Using the equation for vertical motion, we can find the initial velocity (u) of the rock:
u = v - g * t
Where:
v = final velocity (zero at maximum height)
g = acceleration due to gravity (32 ft/sec^2)
t = time (2 seconds)
Substituting the given values into the equation:
u = 0 - (32 ft/sec^2) * 2 sec = -64 ft/sec
Therefore, the initial velocity of the rock is -64 ft/sec (upward).
(b) To find the height of the cliff, we need to calculate the distance traveled by the rock from the maximum height to the ground. We know that it takes 5 seconds for the rock to hit the ground after reaching its maximum height, totaling 7 seconds of time.
Using the equation for vertical motion:
s = u * t + (1/2) * g * t^2
Where:
s = distance traveled
u = initial velocity (-64 ft/sec)
t = time (7 seconds)
g = acceleration due to gravity (32 ft/sec^2)
Substituting the given values into the equation:
s = (-64 ft/sec) * 7 sec + (1/2) * (32 ft/sec^2) * (7 sec)^2
= -448 ft + (1/2) * (32 ft/sec^2) * 49 sec^2
= -448 ft + 16 ft/sec^2 * 49 sec^2
= -448 ft + 784 ft
= 336 ft
Therefore, the height of the cliff is 336 ft.
(c) To find the velocity of the rock at time t (the height of the cliff), we can use the initial velocity and the acceleration due to gravity:
v = u + g * t
Where:
v = velocity
u = initial velocity (-64 ft/sec)
g = acceleration due to gravity (32 ft/sec^2)
t = time (2 seconds)
Substituting the given values into the equation:
v = (-64 ft/sec) + (32 ft/sec^2) * 2 sec
= -64 ft/sec + 64 ft/sec
= 0 ft/sec
Therefore, the velocity of the rock at time t (the height of the cliff) is 0 ft/sec.
(d) The velocity of the rock when it hits the ground can be found using the equation for vertical motion:
v = u + g * t
Where:
v = velocity
u = initial velocity (-64 ft/sec)
g = acceleration due to gravity (32 ft/sec^2)
t = time (7 seconds)
Substituting the given values into the equation:
v = (-64 ft/sec) + (32 ft/sec^2) * 7 sec
= -64 ft/sec + 224 ft/sec
= 160 ft/sec
Therefore, the velocity of the rock when it hits the ground is 160 ft/sec.
To solve this problem, we can use the equations of motion for objects in free fall.
Let's break down the problem step by step:
(a) Initial velocity of the rock:
To find the initial velocity of the rock, we need to know the time it takes for the rock to reach its maximum height. From the given information, we know that it takes 2 seconds.
The velocity of an object at its maximum height is zero, as it momentarily stops before falling back down. Therefore, we can use the equation for vertical velocity:
v = u + gt
Where:
v = final velocity (zero in this case)
u = initial velocity
g = acceleration due to gravity (approximately 32 ft/sec^2)
t = time (2 seconds in this case)
Rearranging the equation, we can solve for u:
0 = u + (32 ft/sec^2) * (2 sec)
Simplifying the equation:
0 = u + 64 ft/sec^2
Rearranging again to solve for u:
u = -64 ft/sec^2
The negative sign indicates that the initial velocity is in the opposite direction of the acceleration due to gravity. So, the initial velocity of the rock is 64 ft/sec in the downward direction.
(b) Height of the cliff:
To find the height of the cliff, we need to know the time it takes for the rock to hit the ground after being thrown.
From the given information, we know that the rock hits the ground 7 seconds (2 seconds to reach maximum height + 5 seconds after that) after it is thrown.
We can use the equation for vertical distance traveled:
s = ut + 0.5 * g * t^2
Where:
s = vertical distance traveled (height of the cliff)
u = initial velocity (-64 ft/sec)
g = acceleration due to gravity (approximately 32 ft/sec^2)
t = time (7 seconds)
Plugging in the values, we can solve for s:
s = (-64 ft/sec) * (7 sec) + 0.5 * (32 ft/sec^2) * (7 sec)^2
Simplifying:
s = -448 ft + 784 ft
s = 336 ft
Therefore, the height of the cliff is 336 ft.
(c) Velocity of the rock at time t:
At any time t, we can find the velocity of the rock using the equation:
v = u + gt
Where:
v = final velocity at time t
u = initial velocity (-64 ft/sec)
g = acceleration due to gravity (approximately 32 ft/sec^2)
t = time
Plugging in the values, we can solve for v at time t:
v = (-64 ft/sec) + (32 ft/sec^2) * t
(d) Velocity of the rock when it hits the ground:
To find the velocity of the rock when it hits the ground, we can substitute the time it takes for the rock to hit the ground (7 seconds) into the equation we used in part (c):
v = (-64 ft/sec) + (32 ft/sec^2) * (7 sec)
Simplifying the equation:
v = (-64 ft/sec) + (224 ft/sec)
v = 160 ft/sec
Therefore, the velocity of the rock when it hits the ground is 160 ft/sec.