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Welcome to the article 1 During a fireworks display a shell is shot into the air with an initial speed of 80 0 m s at an angle of. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!

1. During a fireworks display, a shell is shot into the air with an initial speed of 80.0 m/s at an angle of 85.0° above the horizon. The fuse is timed to ignite the shell just as it reaches its highest point above the ground.

(a) Calculate the height at which the shell explodes.
(b) How much time passed between the launch of the shell and the explosion?
(c) What is the horizontal displacement of the shell when it explodes?

Answer :

The height at which the shell explodes is 233.48 meters, the time between the launch and the explosion is 6.91 seconds, and the horizontal displacement when the shell explodes is 125.32 meters.

Physics of Fireworks Projectile

To calculate the height at which the shell explodes (maximum height), we use the formula for the vertical motion under gravity, where the initial vertical velocity is given by vy = v0sin(θ) and the acceleration is due to gravity (g = 9.8 m/s2, downwards).

(a) Calculating the Height at Which the Shell Explodes

Using the initial vertical velocity of 70.0 m/s and the angle of 75.0°, vy = 70sin(75°) = 67.68 m/s. The shell will reach its highest point when its vertical velocity is zero. At this point, the vertical displacement (y) is the maximum height:

0 = vy2 - 2gy

0 = (67.68 m/s)2 - 2(9.8 m/s2)y

y = (67.68 m/s)2 / (2 x 9.8 m/s2)

y = 233.48 m

(b) Time Passed Between Launch and Explosion

The time to reach the highest point is the time taken for the vertical velocity to become zero due to gravity, calculated using t = vy / g:

t = 67.68 m/s / 9.8 m/s2

t = 6.91 s

(c) Horizontal Displacement of the Shell When it Explodes

Horizontal displacement (horizontal range) is calculated by multiplying the horizontal velocity with the time in the air. The initial horizontal velocity is vx = v0cos(θ) = 70cos(75°) = 18.13 m/s. The horizontal displacement is:

x = vx x t

x = 18.13 m/s x 6.91 s

x = 125.32 m

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Rewritten by : Brahmana