An artilery shell is fired with un Initial velocity of 300 m/s at 45.0° above the horizontal. To clear an avalanche, It explodes on a mountalnside 41.5 s after firing. What are the x- and y-coordinates of the shel where it explodes, relative to its firing point? *************............ ... ..... *.................... 3. Subm Arawet

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**Projectile Motion Problem**

An artillery shell is fired with an initial velocity of 300 m/s at an angle of 45.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 41.5 seconds after firing. What are the x- and y-coordinates of the shell when it explodes, relative to its firing point?

**x =** ___ m  
**y =** ___ m  

*[Submit Answer]*

This problem involves the application of kinematic equations to determine the coordinates of a projectile at a given time. Typically, you would use the initial velocity, angle, and time to calculate both the horizontal (x) and vertical (y) positions using the following formulas:

- \( x = v_0 \cdot t \cdot \cos(\theta) \)

- \( y = v_0 \cdot t \cdot \sin(\theta) - \frac{1}{2} g t^2 \)

Where:
- \( v_0 \) is the initial velocity (300 m/s in this case).
- \( \theta \) is the launch angle (45.0°).
- \( t \) is the time (41.5 s).
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

Solve these equations to find the precise x and y coordinates at the moment of explosion.
Transcribed Image Text:**Projectile Motion Problem** An artillery shell is fired with an initial velocity of 300 m/s at an angle of 45.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 41.5 seconds after firing. What are the x- and y-coordinates of the shell when it explodes, relative to its firing point? **x =** ___ m **y =** ___ m *[Submit Answer]* This problem involves the application of kinematic equations to determine the coordinates of a projectile at a given time. Typically, you would use the initial velocity, angle, and time to calculate both the horizontal (x) and vertical (y) positions using the following formulas: - \( x = v_0 \cdot t \cdot \cos(\theta) \) - \( y = v_0 \cdot t \cdot \sin(\theta) - \frac{1}{2} g t^2 \) Where: - \( v_0 \) is the initial velocity (300 m/s in this case). - \( \theta \) is the launch angle (45.0°). - \( t \) is the time (41.5 s). - \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). Solve these equations to find the precise x and y coordinates at the moment of explosion.
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