Physics, 31.07.2019 22:30 leo4687

Aplane, diving with constant speed at an angle of 40.9â° with the vertical, releases a projectile at an altitude of 654 m. the projectile hits the ground 6.70 s after release. (a) what is the speed of the plane? (b) how far does the projectile travel horizontally during its flight? what were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (state your answers to (c) and (d) as positive numbers.)

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Answer from: scalderon2001

Let's solve the problem step-by-step.

A) Let's start writing the laws of motion on both x (horizontal) and y (vertical) axis. On the x-axis, it's a uniform motion with constant velocity, while on the y-axis it's an uniformly accelerated motion:
S_x(t)=v_x t

$S_y(t)=h-v_y t- \frac{1}{2} g t^2$
where h=654 m is the initial height of the projectile, v_x

and

are the initial velocities on the x- and y-axis, g=9.81 m/s^2 and t is the time. v_y

have a negative signs because they both point downward.
We can calculate the initial velocity on the y-axis by requiring that S_y(6.70 s)=0

, since we know that after 6.70 s the projectile reached the ground. Therefore:
$0=h-v_yt- \frac{1}{2} gt^2=654 - 6.7 v_y - \frac{1}{2} (9.81)(6.70)^2$
from which we find
v_y=64.78 m/s

Then we can find the magnitude of the initial velocity v using the angle with respect to the vertical $\alpha=40.9^{\circ}$ :
$v_y = v cos \alpha$
$v= \frac{v_y}{cos \alpha} =85.7 m/s$

B) Let's calculate the component of the initial velocity on the x-axis:
$v_x = v sin \alpha = 85.7 m/s \cdot sin (40.9^{\circ})=56.11 m/s$
And then, we can find how far the projectile traveled horizontally by calculating Sx at t=6.70 s, when it hits the ground:
$S_x(6.70 s)=v_x t = 56.11 m/s \cdot 6.7 s=375.94 m$

C) The horizontal component of the velocity does not change during the motion, since it's an uniform motion on the x-axis. Therefore, vx at t=6.7 s is the same as its initial value:
v_x (6.70 s)=v_x = 56.11 m/s

D) Instead, vy changes during the motion since it's an accelerated motion, following the law
v_y(t) = - v_y -gt

Using t=6.7 s, we can find the vertical velocity just before the projectile hits the ground
v_y(6.70 s)= -64.78 m/s - (9.81 m/s^2)(6.70 s)=-130.51 m/s

Writing it with positive sign:
v_y(6.70 s) = 130.51 m/s

Answer from: Quest

it is called kinetic energy

Answer from: Quest

distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

slope of a graph = change in value of y -axis / change in values of x -axis

comparing both the equations

change in value of y -axis = change in position

change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

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