High School

Welcome to the article You throw a bomb from the ground with a speed of 8 9 m s at a 50 angle from the horizontal A timer can. 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!

You throw a bomb from the ground with a speed of 8.9 m/s at a 50° angle from the horizontal. A timer can be set so that the bomb explodes when the timer reaches zero. The bomb will also explode if it hits the ground. The bomb's explosion will affect everything in a 6.0 m radius.

What should you set the timer to so that the bomb explodes at its maximum height?

Answer :

Set the timer to approximately 0.348 seconds to ensure the bomb explodes at its maximum height.

Given:

Initial speed (vi) = 8.9 m/s

Launch angle (θ) = 50 degrees

Radius of explosion (r) = 6.0 m

Acceleration due to gravity (g) ≈ 9.8 m/s²

Step 1: Calculate Initial Vertical Velocity (viy):

viy = vi * sin(θ)

viy = 8.9 m/s * sin(50 degrees)

viy ≈ 8.9 m/s * 0.766

viy ≈ 6.81 m/s

Step 2: Calculate Time to Maximum Height (t):

t = viy / g

t = 6.81 m/s / 9.8 m/s²

t ≈ 0.696 s

Step 3: Calculate Timer Setting:

Since we want the bomb to explode at the maximum height, the timer setting is half of the time of flight.

Timer setting = t / 2

Timer setting = 0.696 s / 2

Timer setting ≈ 0.348 s

Final Answer:

Set the timer to approximately 0.348 seconds to ensure the bomb explodes at its maximum height.

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Rewritten by : Brahmana