A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m/s at an angle of 35.0 ∘ with the horizontal.
A). Determine the time taken by the projectile to hit point P at ground level. t= s
B). Determine the distance X of point P from the base of the vertical cliff. Deltax= m
C). At the instant just before the projectile hits point P, determine the horizontal and the vertical components of its velocity.
Enter your answers numerically separated by a comma. Vx, Vy= , m/s
D). At the instant just before the projectile hits point P, determine the magnitude of the velocity. v= m/s
E). At the instant just before the projectile hits point P, determine the angle made by the velocity vector with the horizontal. Theta= degrees below the horizon.
The correct answer and explanation is :
To solve this projectile motion problem, let’s break it down step by step. Given:
- Initial height, ( y_0 = 115 ) m
- Initial speed, ( v_0 = 65.0 ) m/s
- Launch angle, ( \theta_0 = 35.0^\circ )
- Acceleration due to gravity, ( g = 9.81 ) m/s²
A) Time taken to hit the ground
We first determine the total time of flight using the vertical motion equation:
[
y = y_0 + v_{0y} t – \frac{1}{2} g t^2
]
where
( v_{0y} = v_0 \sin\theta_0 = 65.0 \sin 35^\circ = 37.24 ) m/s.
At impact, ( y = 0 ):
[
0 = 115 + (37.24)t – \frac{1}{2} (9.81)t^2
]
Solving this quadratic equation:
[
4.905t^2 – 37.24t – 115 = 0
]
Using the quadratic formula:
[
t = \frac{-(-37.24) \pm \sqrt{(-37.24)^2 – 4(4.905)(-115)}}{2(4.905)}
]
[
t = \frac{37.24 \pm \sqrt{1387.4 + 2254.3}}{9.81}
]
[
t = \frac{37.24 \pm \sqrt{3641.7}}{9.81}
]
[
t = \frac{37.24 \pm 60.36}{9.81}
]
Taking the positive root:
[
t = \frac{37.24 + 60.36}{9.81} = \frac{97.6}{9.81} = 9.95 \text{ s}
]
B) Horizontal Distance (Range)
The horizontal component of velocity remains constant:
[
v_{0x} = v_0 \cos\theta_0 = 65.0 \cos 35^\circ = 53.27 \text{ m/s}
]
[
\Delta x = v_{0x} \cdot t = 53.27 \times 9.95 = 530.1 \text{ m}
]
C) Velocity Components Just Before Impact
The horizontal velocity remains constant:
[
V_x = 53.27 \text{ m/s}
]
For vertical velocity at impact:
[
V_y = v_{0y} – g t = 37.24 – (9.81 \times 9.95)
]
[
V_y = 37.24 – 97.65 = -60.41 \text{ m/s}
]
(negative sign indicates downward direction)
Thus, ( V_x, V_y = 53.27, -60.41 ) m/s.
D) Magnitude of Velocity
[
V = \sqrt{V_x^2 + V_y^2}
]
[
V = \sqrt{(53.27)^2 + (-60.41)^2}
]
[
V = \sqrt{2837.7 + 3649.1} = \sqrt{6486.8} = 80.55 \text{ m/s}
]
E) Angle with Horizontal
[
\theta = \tan^{-1} \left( \frac{|V_y|}{V_x} \right) = \tan^{-1} \left( \frac{60.41}{53.27} \right)
]
[
\theta = \tan^{-1} (1.134) = 48.2^\circ
]
Final Answers:
A) t = 9.95 s
B) Δx = 530.1 m
C) ( V_x, V_y = 53.27, -60.41 ) m/s
D) ( V = 80.55 ) m/s
E) ( \theta = 48.2^\circ ) below the horizon