Problem with projectile velocity

In summary, the ball is thrown leftward from the left edge of the roof of a building at height h above the ground. The ball hits the ground 1.50s later, at distance d=25.0m from the building and at angle theta = 60 degrees with the horizontal. The ball has an initial velocity of 16.67m/s (25 meters/1.5 seconds) and hits the ground with a vertical velocity of -28.87m/s. Using this information, the height of the ball can be found to be 32.3m.
  • #1
sabres
4
0
A ball is thrown leftward from the left edge of the roof of a building at height h above the ground. The ball hits the ground 1.50s later, at distance d=25.0m from the building and at angle theta = 60 degrees with the horizontal.

A) Find h.

What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown?



I figured that initial velocity in X direction is 16.67 m/s (25 meters/1.5 seconds). What is really giving me trouble is finding the magnitude of the velocity. I can't figure out how to find the Y component of the velocity to get it. I'm not even sure what B and C want, but I figure i'll cross that bridge after I get A handled.
 
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  • #2
Use the angle given 60 degrees, to find the height.
 
  • #3
Could you be more specific please? I don't know how to do that.
 
  • #4
If I am thinking of this right then your angle is opposite of your horizontal so you use tan(60)x25.0 m to find the horizontal.
 
  • #5
Fizicks said:
If I am thinking of this right then your angle is opposite of your horizontal so you use tan(60)x25.0 m to find the horizontal.

yes, tan(60)*25 is the height h.
 
  • #6
That is what I thought originally, too. Problem is the 25 is not the hypotenuse of the triangle that is formed, so (tan 60) * 25 yields an incorrect answer. That gives 43.3, but the back of the book says it is 32.3
 
  • #7
sabres said:
That is what I thought originally, too. Problem is the 25 is not the hypotenuse of the triangle that is formed, so (tan 60) * 25 yields an incorrect answer. That gives 43.3, but the back of the book says it is 32.3

Ah... looks like I misunderstood the problem. The object is thrown at an angle, so it has an initial vertical velocity... it hits the ground with an angle of 60... that means that (vertical speed)/(horizontal speed) is tan(60) when the object hits.

ie it hits with a vertical velocity of - 16.67tan(60) = -28.87m/s

using this you can get the height, and it comes out to 32.3m.
 
  • #8
You're a lifesaver, thank you!
 

1. What is projectile velocity?

Projectile velocity is the speed at which an object is projected or launched into motion. It is a vector quantity that includes both magnitude (speed) and direction.

2. What factors affect projectile velocity?

There are several factors that can affect projectile velocity, including the angle of projection, air resistance, and the initial velocity of the object. Gravity also plays a significant role in determining the velocity of a projectile.

3. How is projectile velocity calculated?

Projectile velocity can be calculated using the equation v= u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. The velocity can also be determined using trigonometry and the angle of projection.

4. What is the difference between initial and final projectile velocity?

The initial velocity of a projectile is the velocity at which the object is launched or projected into motion. The final velocity is the velocity at which the object lands or stops moving. The difference between the two is affected by factors such as air resistance and the angle of projection.

5. How does air resistance impact projectile velocity?

Air resistance, also known as drag, can significantly impact projectile velocity by slowing down the object as it moves through the air. This is due to the friction between the object and the air molecules. Objects with a larger surface area or irregular shape experience more air resistance, resulting in a decrease in velocity.

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