Projectile Motion Launched at an Angle

In summary, the golf ball is thrown from the roof of a school with a velocity of 20 ms at an angle of 30 deg above the horizontal. The equation for time of flight is D=v1(t)+1/2a(t)^2, and the equation for horizontal displacement is H=v1t+1/2a(t)^2.
  • #1
phizics09
38
1

Homework Statement


A golf ball is launched from the roof of a school with a velocity of 20 ms at an angle of 30 deg above the horizontal.

Homework Equations


If the roof is 40m above the ground, calculate
a) the ball's time of flight
b)the ball's horizontal displacement

The Attempt at a Solution


So for the first question, the example in the book only uses one equation to solve for t, D=v1(t)+1/2a(t)^2,but I thought since the ball goes up and comes down, there should be two separate parts to the solution?
 
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  • #2
No, it's a single parabola. However, we should still break the initial vector they give you into components to separately model x and y positions.

Once you've done that, you can set the equation modeling y position to 0, then find the value of t that makes it true, and you have time of flight.

Given that time for t, you can plug that into the x position equation and you have horizontal displacement.

Do you know what to do with the vector they've given you?
 
  • #3
Yeah, I get it now. So if it wasn't a projectile problem, I would still use the same D for the entire question? For example, if a kinematics problem has a ball thrown upwards at the edge of a the top of a building with an initial vertical velocity, and the ball lands on the ground below, and it asks you to find the time it takes to land on the ground, you would just use the height of the building in the equation D=v1t+1/2a(t)^2 to find t?
 
  • #4
Yes, your equation for the y position of the projectile will only depend on the initial y velocity and the initial position, but nothing to do with the x position/velocity.

You can always break it into components this way. Just remember to do it correctly. Launching it at 20 m/s at a 30 degree angle is the same as launching it at 20sin30 m/s vertically.
 
  • #5
Thanks!
 

1. What is projectile motion launched at an angle?

Projectile motion launched at an angle refers to the motion of an object that is launched horizontally at an angle from the ground. This type of motion is a combination of horizontal motion (constant velocity) and vertical motion (acceleration due to gravity).

2. What factors affect the trajectory of a projectile launched at an angle?

The factors that affect the trajectory of a projectile launched at an angle include the initial velocity, the launch angle, and the force of gravity. Other factors such as air resistance and wind can also have an impact on the trajectory.

3. How does the launch angle impact the range of a projectile?

The launch angle is directly related to the range of a projectile. The range will be maximum when the launch angle is 45 degrees. If the launch angle is increased, the range will decrease and if the launch angle is decreased, the range will increase.

4. What is the maximum height reached by a projectile launched at an angle?

The maximum height reached by a projectile launched at an angle is determined by the initial velocity and the launch angle. The maximum height will be reached when the projectile is at its highest point and has zero vertical velocity. The formula to calculate the maximum height is h = (v02sin2θ) / (2g), where h is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

5. How does air resistance affect the motion of a projectile launched at an angle?

Air resistance, also known as drag, can have a significant impact on the motion of a projectile launched at an angle. It can decrease the range and height of the projectile, and can also change the shape of its trajectory. However, the effect of air resistance is usually small and can be ignored in most cases.

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