Projectile motion - velocity and distance

In summary, the problem involves a projectile launched at 60 m/s at a 30 degree angle, landing on a hillside after 4 seconds. The velocity at the highest point in the trajectory is calculated using the formula for final velocity and the distance from the launch point to the target is calculated using the formula for displacement. The final velocity at the highest point was incorrectly calculated as 10.4 m/s and the distance was calculated as 207.846097 m, but the correct answer is 212 m.
  • #1
piknless
9
0
Projectile motion -- velocity and distance

Homework Statement


A projectile is launched with an initial speed of 60 m/s at an angle of 30 degrees above the horizontal. The projectile lands on a hillside 4 seconds later. Neglect air friction. What is the projectile's velocity at the highest point of it's trajectory? What is the straight line distance from where the projectile launched to where it hits it's target?

Homework Equations


x=vt

The Attempt at a Solution


For my B I got 207 m but the answer in the book says 212 m.
 
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  • #2


piknless said:

Homework Statement


A projectile is launched with an initial speed of 60 m/s at an angle of 30 degrees above the horizontal. The projectile lands on a hillside 4 seconds later. Neglect air friction. What is the projectile's velocity at the highest point of it's trajectory? What is the straight line distance from where the projectile launched to where it hits it's target?


Homework Equations


x=vt


The Attempt at a Solution


For my B I got 207 m but the answer in the book says 212 m.

Can you please show us your work step-by-step? That will make it easier to help you.

Also, I will add a bit to your one-word thread title -- please try to make your thread titles very descriptive of the specific question you are asking in the thread.
 
  • #3


I'm not exactly sure how to use symbols, but my work is as follows. First I split the vector into the x and y components. I got the vector component x to be 51.961523423 and the vector y component to be 30. The I used the formula: final velocity squared equals initial velocity squared plus 2 times change in y time acceleration. The initial velocity is 30 m/s. The change in y is 2 times velocity times 1/2 time plus acceleration times 1/2 times time squared. I get the cahnge in y to be 40.4 I got the final velocity to equal 10.4. That is incorrect. For part b I used x=vt velocity is 51.96152423 and the time is 4 seconds. multiplied I get 207.
 

What is projectile motion?

Projectile motion refers to the motion of an object that is launched or thrown into the air and moves along a curved path under the influence of gravity.

How is velocity calculated for projectile motion?

The velocity of a projectile is calculated by dividing the distance it travels by the time it takes to travel that distance. This can be represented by the equation v = d/t, where v is velocity, d is distance, and t is time.

What factors affect the distance and velocity of a projectile?

The distance and velocity of a projectile are affected by several factors, including the initial velocity, angle of launch, air resistance, and the acceleration due to gravity.

Can the velocity of a projectile change during its flight?

Yes, the velocity of a projectile can change during its flight due to the influence of external forces such as air resistance or wind. However, the acceleration due to gravity remains constant throughout the motion.

How does the angle of launch affect the distance and velocity of a projectile?

The angle of launch plays a crucial role in determining the distance and velocity of a projectile. The optimal angle of launch for maximum distance is 45 degrees, while a lower angle will result in a shorter distance but a higher velocity. A higher angle will result in a longer distance but a lower velocity.

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