Consider a rock that is thrown off a bridge of height 27 m at an angle θ = 24° with respect to the horizontal as shown in the figure above. If the initial speed the rock is thrown is 12 m/s, find the following quantities:
(a) The time it takes the rock to reach its maximum height.
s
(b) The maximum height reached by the rock.
m
(c) The time at which the rock lands.
s
(d) The place where the rock lands.
m
(e) The velocity of the rock (magnitude and direction) just before it lands.
Magnitude Direction
m/s °
a) Let the rock attain its maximum height in t1 sec
=>By v = u - gt
=>0 = [Uy] - gt1
=>t1 = usinθ/g
=>t1 = [12 x sin24*]/9.8
=>t1 ≈ 0.50 sec
(b) Let the rock attain h meter in t1 sec
=>By v^2 = u^2 - 2gh
=>0 = [Uy]^2 - 2gh
=>h = [12 x sin24*]^2/[2 x 9.8]
=>h = 1.22 m
(c) Total height (H) of the stone fro ground = h + 27 = 28.22 m
Let the stone take t2 sec to fall H meter
=>By s = ut + 1/2gt^2
=>28.22 = 0 + 1/2 x 9.8 x t2^2
=>t2 = √5.76
=>t2 ≈ 2.4sec
Thus the total time to land (T) = t1 + t2 = 0.50 + 2.4 = 2.90 sec
(d) By R = [Ux] x T
=>R = ucosθ x T
=>R = 12 x cos24* x 2.90
=>R = 31.79 m
(e) Let the angle of landing is α and the magnitude of velocity is v m/s
=>By v = u + gt
=>Vy = 0 + 9.8 x t2
=>Vy = 9.8 x 2.4 = 23.52 m/s
& Vx = Ux = 10.96 m/s
=>v = √[(Vx)^2 + (Vy)^2]
=>v = √[(10.96)^2 + (23.52)^2]
=>v = 25.95 m/s
By tanα = Vy/Vx = 23.52/10.96 = 2.15 ≈ tan65*
=>α = 65*
B C E are wrong what am I doing wrong?
Let's go through the calculations step by step and see if we can identify any errors:
(a) The time it takes the rock to reach its maximum height:
To find the time it takes for the rock to reach its maximum height, we can use the formula:
t1 = (Vy - Uy) / g
Using the given values:
Uy = u * sin(θ) = 12 * sin(24°)
g = 9.8 m/s^2
Plugging in the values:
t1 = (12 * sin(24°)) / 9.8
t1 ≈ 0.405 sec (rounded to three decimal places)
(b) The maximum height reached by the rock:
To find the maximum height, we can use the formula:
h = Uy * t1 - (1/2) * g * t1^2
Using the known values:
Uy = 12 * sin(24°)
t1 = 0.405 sec
g = 9.8 m/s^2
Plugging in the values:
h = (12 * sin(24°)) * 0.405 - (1/2) * 9.8 * (0.405)^2
h ≈ 1.084 m (rounded to three decimal places)
(c) The time at which the rock lands:
To find the time at which the rock lands, we need to consider the total time it takes for the rock to hit the ground. Since the rock is thrown at an angle, it will reach its maximum height and then fall back down.
The total time is given by:
T = 2 * t1
Using the known value for t1:
T = 2 * 0.405
T ≈ 0.81 sec (rounded to two decimal places)
(d) The place where the rock lands:
To find the horizontal distance traveled by the rock, we can use the formula:
R = Ux * T
Using the given values:
Ux = u * cos(θ) = 12 * cos(24°)
T = 0.81 sec
Plugging in the values:
R = (12 * cos(24°)) * 0.81
R ≈ 9.42 m (rounded to two decimal places)
(e) The velocity of the rock just before it lands:
To find the velocity just before the rock lands, we can break it down into its horizontal and vertical components:
Vx = Ux
Vy = Uy - g * t2
To find t2, the time it takes for the rock to fall from its maximum height to the ground, we can use the formula:
t2 = √(2 * h / g)
Using the known values:
h = 1.084 m
g = 9.8 m/s^2
Plugging in the values:
t2 = √(2 * 1.084 / 9.8)
t2 ≈ 0.464 sec (rounded to three decimal places)
Now we can find Vy and Vx:
Vy = (12 * sin(24°)) - (9.8 * 0.464)
Vy ≈ 4.645 m/s (rounded to three decimal places)
Vx = 12 * cos(24°)
Vx ≈ 10.688 m/s (rounded to three decimal places)
Finally, we can calculate the magnitude and direction of the velocity:
v = √(Vx^2 + Vy^2)
v ≈ √(10.688^2 + 4.645^2)
v ≈ 11.656 m/s (rounded to three decimal places)
tan(α) = Vy / Vx
α = arctan(Vy / Vx)
Plugging in the values:
α ≈ arctan(4.645 / 10.688)
α ≈ 23.773° (rounded to three decimal places)
Therefore, the corrected answers are:
(a) The time it takes the rock to reach its maximum height: approximately 0.405 sec
(b) The maximum height reached by the rock: approximately 1.084 m
(c) The time at which the rock lands: approximately 0.81 sec
(d) The place where the rock lands: approximately 9.42 m
(e) The velocity of the rock just before it lands (magnitude and direction): approximately 11.656 m/s at an angle of approximately 23.773°
In order to find the correct answers for (b), (c), and (e), you need to consider a few things:
(b) The maximum height reached by the rock:
To find the maximum height, you need to compute the vertical component of the initial velocity using the angle θ. The vertical component of the initial velocity can be calculated as:
[Uy] = usinθ = 12 m/s * sin(24°) = 4.67 m/s
Now, you can find the maximum height using the kinematic equation:
v^2 = u^2 - 2gh
Set the final velocity v = 0, initial velocity u = [Uy], and solve for h:
0 = ([Uy])^2 - 2 * 9.8 m/s^2 * h
h = ([Uy])^2 / (2 * 9.8 m/s^2) = (4.67 m/s)^2 / (2 * 9.8 m/s^2) ≈ 1.09 m
Therefore, the correct answer for (b) is approximately 1.09 m.
(c) The time at which the rock lands:
To find the time at which the rock lands, you can use the equation for the total time T:
T = t1 + t2
From part (a), you calculated t1 = 0.50 seconds. Now, you need to find t2, the time it takes for the rock to fall from the maximum height of h.
Using the equation for displacement in the vertical direction:
h = ([Uy]) * t2 - (1/2) * g * (t2)^2
Rearrange the equation to solve for t2:
(t2)^2 - (4.67 m/s) * t2 - (1.09 m) * 2 / (-9.8 m/s^2) = 0
Solve this quadratic equation to find t2, and then calculate T = t1 + t2.
(e) The velocity of the rock just before it lands:
To find the velocity just before the rock lands, you can use the equation:
v = u + gt
The horizontal component of the velocity remains constant throughout the motion, so the horizontal velocity just before landing is the same as the initial horizontal velocity (constant velocity).
The vertical velocity can be found using the equation:
Vy = ([Uy]) + g * t2
Calculate Vy and then use the Pythagorean theorem to find the magnitude of the final velocity v = √((Vx)^2 + (Vy)^2). Finally, use the inverse tangent function to find the direction of the final velocity α.
By recalculating these quantities using the correct equations, you should find the correct answers for (b), (c), and (e).