A projectile shot at an angle of 60 degrees above the horizontal strikes a building 25m away at a point 16m above the point of projection. Find the magnitude and direction of the v... A projectile shot at an angle of 60 degrees above the horizontal strikes a building 25m away at a point 16m above the point of projection. Find the magnitude and direction of the velocity of the projectile as it strikes the building.
Understand the Problem
The question describes a projectile motion scenario where a projectile is fired at a specified angle and strikes a building at a certain distance and height. We need to calculate the magnitude and direction of the projectile's velocity upon impact with the building.
Answer
Magnitude: $11.63 \, m/s$ Direction: $-24.23^\circ$
Answer for screen readers
Magnitude of the velocity at impact: $11.63 , m/s$ Direction of the velocity at impact: $-24.23^\circ$ (below the horizontal)
Steps to Solve
- Resolve the initial velocity into horizontal and vertical components.
Let $v_0$ be the initial velocity. The horizontal component, $v_{0x}$, and vertical component, $v_{0y}$, are given by:
$v_{0x} = v_0 \cos(60^\circ) = \frac{1}{2} v_0$
$v_{0y} = v_0 \sin(60^\circ) = \frac{\sqrt{3}}{2} v_0$
- Use the horizontal motion to find the time of impact.
Since there is no horizontal acceleration, the horizontal velocity remains constant. We can use the horizontal distance $x = 25$ m to find the time $t$ it takes to reach the building:
$x = v_{0x} t$ $25 = \frac{1}{2} v_0 t$ $v_0 t = 50$ (Equation 1)
- Use the vertical motion to relate time and initial velocity.
The vertical motion is affected by gravity ($g = 9.8 , m/s^2$). We can use the vertical displacement $y = 16$ m to relate the initial vertical velocity and time:
$y = v_{0y} t - \frac{1}{2} g t^2$ $16 = \frac{\sqrt{3}}{2} v_0 t - \frac{1}{2} (9.8) t^2$ $16 = \frac{\sqrt{3}}{2} (50) - 4.9 t^2$ (Substitute $v_0 t = 50$ from Equation 1) $16 = 25\sqrt{3} - 4.9 t^2$ $4.9 t^2 = 25\sqrt{3} - 16$ $4.9 t^2 = 43.30 - 16$ $4.9 t^2 = 27.30$ $t^2 = \frac{27.30}{4.9} \approx 5.57$ $t \approx \sqrt{5.57} \approx 2.36 , s$
- Calculate the initial velocity $v_0$.
Using the time $t$ and Equation 1: $v_0 t = 50$ $v_0 = \frac{50}{t} = \frac{50}{2.36} \approx 21.19 , m/s$
- Calculate the horizontal and vertical components of the velocity at impact.
The horizontal component $v_x$ remains constant: $v_x = v_{0x} = \frac{1}{2} v_0 = \frac{1}{2} (21.19) \approx 10.60 , m/s$
The vertical component $v_y$ changes due to gravity: $v_y = v_{0y} - gt = \frac{\sqrt{3}}{2} v_0 - gt = \frac{\sqrt{3}}{2} (21.19) - (9.8)(2.36)$ $v_y \approx 18.34 - 23.13 \approx -4.79 , m/s$
- Calculate the magnitude of the velocity at impact.
The magnitude of the velocity $v$ is: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10.60)^2 + (-4.79)^2} \approx \sqrt{112.36 + 22.94} = \sqrt{135.30} \approx 11.63 , m/s$
- Calculate the direction of the velocity at impact.
The angle $\theta$ of the velocity with respect to the horizontal is: $\theta = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{-4.79}{10.60}\right) \approx \arctan(-0.45) \approx -24.23^\circ$
Magnitude of the velocity at impact: $11.63 , m/s$ Direction of the velocity at impact: $-24.23^\circ$ (below the horizontal)
More Information
The negative angle indicates that the projectile is traveling downwards when it hits the building.
Tips
- Incorrectly resolving initial velocity components: Ensure correct trigonometric functions are used, sine for vertical and cosine for horizontal components, relative to the launch angle.
- Forgetting the effect of gravity on vertical motion: The constant acceleration due to gravity must be included in the vertical motion equation.
- Using incorrect signs: Pay close attention to the signs of velocity and acceleration components in each direction ($g$ is usually negative in this coordinate system).
- Mixing up horizontal and vertical motion: Remember that the horizontal motion has constant velocity, while vertical motion is influenced by gravity.
- Rounding errors: Avoid rounding intermediate values too early, as this can affect the final answer.
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