The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall?

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Now, onto the question at hand. We're dealing with projectile motion here. Given that the ceiling of the hall is 25 meters high and the initial speed of the ball is 40 m/s, we need to determine the maximum horizontal distance it can travel before hitting the ceiling.

To solve this, we'll use the equations of motion for projectile motion. The key here is to find the time it takes for the ball to reach its maximum height, and then calculate the horizontal distance it travels during that time.

The formula for the time taken to reach maximum height in projectile motion is:

t=vygt=gvy

where:

• vyvy is the vertical component of initial velocity (in this case, the vertical component is the initial velocity itself, as the ball is thrown straight up),
• gg is the acceleration due to gravity (approximately 9.8 m/s29.8m/s2).

Substituting the given values, we find:

t=40 m/s9.8 m/s2≈4.08 st=9.8m/s240m/s≈4.08s

Now, the time taken to reach maximum height is the same as the time taken for the ball to fall back to the ground from maximum height. Therefore, the total time of flight (TT) is twice this value, which is approximately 8.16 s8.16s.

Now, we can find the horizontal distance (dd) using the formula:

d=vx×Td=vx×T

where:

• vxvx is the horizontal component of initial velocity (which remains constant throughout the motion),
• TT is the total time of flight.

The horizontal component of initial velocity (vxvx) is simply the initial velocity itself, as there is no horizontal acceleration. So, vx=40 m/svx=40m/s.

d=40 m/s×8.16 sd=40m/s×8.16s

d≈326.4 md≈326.4m

So, the maximum horizontal distance the ball can travel without hitting the ceiling is approximately 326.4 m326.4m.