Question

An artillery shell is fired with an initial velocity of 331.5m/s at 64.0° above the horizontal. It explodes on a mountainside 35.5s after firing. What is the vertical coordinate (y) of the point where the shell explodes, relative to its firing point? State your answer to the nearest meter. Use g=9.8m/s².

Answer 1

The vertical coordinate (y) of the point where the shell explodes is approximately 5224m (rounded to the nearest meter).

Step-by-step explanation:

The vertical coordinate (y) of the point where the shell explodes can be calculated using the equation for vertical motion:

y = yo + voy×t - (1/2)gt^2

Where:

• yo is the initial vertical position (0m)
• voy is the initial vertical velocity (calculated as vi×sin(θ))
• t is the time of flight (35.5s)
• g is the acceleration due to gravity (9.8m/s²)

Plugging in the given values:

• vi = 331.5m/s
• θ = 64.0°
• t = 35.5s
• g = 9.8m/s²

We can calculate:

• voy = 331.5m/s × sin(64.0°) = 290.77m/s
• y = 0m + 290.77m/s × 35.5s - (1/2) × 9.8m/s² × (35.5s)^2 = 5223.58m

Therefore, the vertical coordinate (y) of the point where the shell explodes is approximately 5224m (rounded to the nearest meter).