Height, angles, and projectiles

[FrenchAtticus]

Homework Statement

A spring-loaded gun can fire a projectile to a height h if it is fired straight up.
If the same gun is pointed at an angle of 45 degrees from the vertical, what maximum height can now be reached by the projectile.

1. h _
2. h / 2\/2
3. h / 2 _
4. h / \/2
5. h / 4

2. The attempt at a solution

I believe it would be h / 2, but I'm not sure why other than the fact that 90/2=45 and the heights would therefore be relative.

 

[tiny-tim]

Welcome to PF!

A spring-loaded gun can fire a projectile to a height h if it is fired straight up.
If the same gun is pointed at an angle of 45 degrees from the vertical, what maximum height can now be reached by the projectile.

Hi FrenchAtticus! Welcome to PF!

Hint: use conservation of energy, and Newton's second law horizontally.

 

[thomate1]

Can someone explain the answer and why it is correct?I can provide an explanation for the correct answer. When a projectile is fired straight up, it follows a parabolic path and reaches a maximum height at the apex of its trajectory. However, when the same projectile is fired at an angle of 45 degrees, it follows a different trajectory known as a projectile motion. In this case, the maximum height reached by the projectile will be less than the height reached when fired straight up. This is because the horizontal component of the velocity of the projectile will contribute to its motion, causing it to cover less vertical distance and therefore reach a lower maximum height.

To calculate the maximum height reached by the projectile when fired at an angle, we can use the equation h = u^2 sin^2(theta) / 2g, where u is the initial velocity, theta is the angle of projection, and g is the acceleration due to gravity. In this case, since we are firing at an angle of 45 degrees, sin^2(theta) = 1/2. Therefore, the maximum height reached will be h / 2, as shown in option 2.

In summary, when a projectile is fired at an angle, its maximum height will be less than when fired straight up due to the horizontal component of its velocity. The correct answer is h / 2, which can be calculated using the equation h = u^2 sin^2(theta) / 2g.