couple of physics question

[smileurdentist]

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An airplane with a speed of 79.6 m/s is climbing upward at an angle of 44.8 ° with respect to the horizontal. When the plane's altitude is 961 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.


A skateboarder shoots off a ramp with a velocity of 7.7 m/s, directed at an angle of 57° above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
I got the (b) which was 2.76 m correct
but (a) was 4.32 and it is wrong!


Please help ASAP!

Thanks in advance

 

[Saint Fu]

Haven't actually crunched any numbers myself, but did you remember to account for the initial 1.4m height?

 

[smileurdentist]

Haven't actually crunched any numbers myself, but did you remember to account for the initial 1.4m height?

I think so... I am super confused!!!!

 

[MI Achilles]

(a) was 4.32 and it is wrong!


Please help ASAP!

Thanks in advance

Wouldn't it be 5.65...
What's your equation for velocity of angles?

 

[Saint Fu]

Wouldn't it be 5.65...
What's your equation for velocity of angles?

That's funny, I'm getting about 3.5. Anyway yeah, just show us how you came to that answer.

The first question should also be very similar to the second. You basically just have a different starting point in the object's trajectory.

 

[circulus vitios]

It's been two years since I've taken physics, but I think this is right:

A. Find the y component of velocity. See (1).

Find the maximum height of the skateboarder. You don't know the time so the only kinematics equation that will work is (2). Substitute the values and get (3).

B. Find the horizontal distance traveled. To do this you need to find the skateboarder's hang time (the time he is in the air). Use (4) and get (5).

Find the horizontal component of velocity. See (6). Find the horizontal distance traveled by (7).

 

[neurodoc]

These are simple kimematic questions that involve simple assumptions. In both cases (airplane dropping a package and skateboarder launching himself into free fall off an inclined plane) and you are given the initial conditions (speed and angle with respect to the horizontal; height above horizontal (ground) where free fall begins for both the package and the skateboarder) and you are asked to calculate in the first case how far along the horizontal from the point of release the package reaches the ground, and in the second case what is the horizontal point on the ground that the skateboarder reaches his highest point after his take-off from the ramp. You solve these two problems by simple vector decomposition into independent horizontal and vertical motions. The force in the y-direction is gravity, with an acceleration of 9.8m/s in the downward direction. The horizontal motion is characterized by the velocity (speed and angle to the horizontal) at the time when the airplane package is released or the skateboarder lauches himself into free fall. 😴