Consider a rock thrown off a bridge of height 78.5 m at an angle è = 25° with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 15.6 m/s. Find the following quantities:
(a) the maximum height reached by the rock
(b) the time it takes the rock to reach its maximum height
(c) the place where the rock lands
(d) the time at which the rock lands
(e) the velocity of the rock (magnitude and direction) just before it lands.
magnitude
direction: Give your answer in degrees
u = horizontal speed = 15.6 cos 25 for the whole problem
Vi = initial speed up = 15.6 sin 25
in general
v = Vi = 9.8 t
h = 78 + Vi t - 4.9 t^2
for max height
v = 0 at top = Vi - 9.8 * time at top
height at top = 78 + Vi t - 4.9 t^2
A)2
b)5.7
C)87.7
D) 76.4
E)4.5
To solve this problem, we can use the kinematic equations of motion. Let's break down each part of the problem and find the answers step by step.
(a) To find the maximum height reached by the rock, we need to find the vertical component of the initial velocity.
Vertical component of initial velocity (Vy) = initial velocity (Vi) × sin(angle è)
Given: Vi = 15.6 m/s, angle è = 25°
Vy = 15.6 m/s × sin(25°)
Vy ≈ 6.5 m/s
Now, we can use the kinematic equation for vertical motion to find the maximum height (H):
H = (Vy^2) / (2 * g)
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
H = (6.5 m/s)^2 / (2 * 9.8 m/s^2)
H ≈ 2.17 m
Therefore, the maximum height reached by the rock is approximately 2.17 meters.
(b) To find the time taken by the rock to reach its maximum height, we can use the kinematic equation for vertical motion:
Vy = Viy - g * t
where Viy is the vertical component of the initial velocity, and t is the time.
Given: Viy = 6.5 m/s, g = 9.8 m/s^2
0 = 6.5 m/s - 9.8 m/s^2 * t
Solving for t:
t = 6.5 m/s / (9.8 m/s^2)
t ≈ 0.66 s
Therefore, it takes approximately 0.66 seconds for the rock to reach its maximum height.
(c) To find the place where the rock lands, we need to find the horizontal distance traveled by the rock.
Horizontal component of initial velocity (Vx) = initial velocity (Vi) × cos(angle è)
Given: Vi = 15.6 m/s, angle è = 25°
Vx = 15.6 m/s × cos(25°)
Vx ≈ 14.13 m/s
Now, we can use the kinematic equation for horizontal motion to find the horizontal distance (D):
D = Vx * t
where t is the time.
Given: t ≈ 0.66 s, Vx ≈ 14.13 m/s
D = 14.13 m/s * 0.66 s
D ≈ 9.31 m
Therefore, the rock lands at a horizontal distance of approximately 9.31 meters.
(d) To find the time at which the rock lands, we can use the same horizontal distance formula as in part (c):
D = Vx * t
Given: D ≈ 9.31 m, Vx ≈ 14.13 m/s
9.31 m = 14.13 m/s * t
Solving for t:
t = 9.31 m / 14.13 m/s
t ≈ 0.66 s
Therefore, the rock lands at approximately 0.66 seconds.
(e) To find the velocity (magnitude and direction) of the rock just before it lands, we can find the vertical and horizontal components of the velocity.
Vertical component of final velocity (Vyf) = Vy - g * t
where Vy is the vertical component of the initial velocity and t is the time.
Given: Vy = 6.5 m/s, g = 9.8 m/s^2 and t ≈ 0.66 s
Vyf = 6.5 m/s - 9.8 m/s^2 * 0.66 s
Vyf ≈ 0 m/s
Horizontal component of final velocity (Vxf) remains the same as Vx, as there is no horizontal acceleration.
Therefore, the velocity of the rock just before it lands is approximately 0 m/s vertically and 14.13 m/s horizontally.
To find the magnitude of the velocity, we can use the Pythagorean theorem:
Magnitude of velocity (V) = √(Vx^2 + Vy^2)
V = √((14.13 m/s)^2 + (0 m/s)^2)
V ≈ 14.13 m/s
The magnitude of the velocity just before the rock lands is approximately 14.13 m/s.
To find the direction, we can use the inverse tangent function:
Direction = arctan(Vyf / Vxf)
Given: Vyf ≈ 0 m/s and Vxf ≈ 14.13 m/s
Direction = arctan(0 / 14.13)
Direction ≈ 0°
Therefore, the direction of the velocity just before the rock lands is approximately 0° with respect to the horizontal.