Problem

Multiple-Concept Example 4 deals with a situation…

02:27

Problem 37 Hard Difficulty

SSM An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0 with respect to the horizontal. When the plane’s altitude is 732 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.

Answer

(a) $x=1381 \mathrm{m}$
(b) $\theta=-66.04^{\circ}$

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Physics 101 Mechanics

Discussion

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Sam H.

August 23, 2021

Why is the 9.81 used to find the time of the vertical component negative ? I thought if an object was travelling downwards with gravity that the gravitational constant would remain constant

Sam H.

August 23, 2021

Sam H.

August 23, 2021

Remain positive not constant

Jennifer L.

September 26, 2021

At the end when trying to find the angle using tan inverse, which V of x are you using??

Alyssa J.

October 16, 2021

to find the angle of impact, use the initial velocity of x because Ax=0 m/s^2

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Video Transcript

So the question states that a projectile is launched from a plane an angle of 50 degrees at a velocity of 97 0.5 meters per second, and this projectiles launched at a height of 732 meters. And our task is to find how far in the X direction the box travels before it lands on the ground as well as the angle of the velocity vector. Once the box reaches the ground. So the first thing we need to do is figure out how long the boxes in the air. Four. And to do this we can separate the velocity vector into components so we can figure out what the horizontal glass defector is because we know co sign of 50 degrees is equal to adjacent overhype on news which is the eggs over 97.5, which means V of X is equal to 97.5 times co sign 15 degrees and the same goes for view of why. And we just change the sign her the coastline to assign. So you guys 97.5 time sign of 50 degrees Now that we know via of y N v of X. We can use arcamax formulas toe so for the time it takes for the box to reach the ground. So in this case, the best formula to use would be the equation that states that the change in displacement is equal to the initial velocity times the time plus one have times the acceleration times that time squared. The change in displacement is negative negative 732 meters. In this case, because it starts at this height, the initial velocity in the vertical direction is 97.5 times sign of 50 degrees. Multiply this times time, then we add 1/2 times the acceleration which in this case is negative. This is supposed to be a negative 9.8 times time squared and to solve for the time, we can either use a calculator or quadratic formula. In this case, I used a calculator and we find that the roots are 22.0 to 5 for the time is a tes equals 22.25 seconds. Now that we know the time, we can just plug it into the formula which states that the change displacement is equal to the initial lossy times the time and this is the initial Aussie in the horizontal direction, which is given by 97.5 times co sign of 50 degrees. Multiply this by 22 0.25 seconds and this gives us a total delta X of 1381 meters. This is this value. Now if we want to find the angle at which the velocity vectors pointing at the time that the box hits the ground, we know for a fact that the horizontal velocity is going to stay constant throughout its motion. So it will be 97.5 co sign of 50 degrees for the horizontal velocity. And if we want to find the vertical velocity at this point, we can use the chemical equation which states that the velocity at Time T is equal to the initial velocity, plus the acceleration times that time. We're trying to find V in this case, this is really visa by, um we know that the initial vertical velocity is 97.5 times sign of 50 degrees and we know that the acceleration is going to be negative. 9.8 and we also know the time which we calculated up above, which is 22.25 seconds. When we work this out, we find that via Y is equal to Ah negative 141.155 meters per second. And now that we know what our via vexes and R V A Y is when the box hits the ground, we can just take the 10 inverse of V of y over the events and that I'll give us our angle here that we want. When we do this, we end up getting negative 66 0.4 degrees.

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