How Fast and Far Does a Projectile Travel When Released from a Diving Plane?

AI Thread Summary
A plane diving at a 51.0° angle releases a projectile from 570 m altitude, which hits the ground in 4.00 seconds. The projectile's motion is influenced solely by gravity, resulting in constant horizontal velocity and vertical acceleration. To determine the aircraft's speed, the equations of motion are applied, factoring in the initial velocity components derived from trigonometric functions. The horizontal distance traveled by the projectile is 570 m, and the vertical impact speed can be calculated using the known time and initial velocities. Understanding the reversed trig functions is crucial for solving the vertical component of the projectile's velocity before impact.
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Homework Statement



A plane, diving with constant speed at an angle of 51.0° with the vertical, releases a projectile at an altitude of 570 m. The projectile hits the ground 4.00 s after release. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.)
(a) What is the speed of the aircraft?


(b) How far did the projectile travel horizontally during its flight?


(c) What were the horizontal and vertical components of its velocity just before striking the ground?



Homework Equations





The Attempt at a Solution


ive tried the vertical and horizontal motion equations but cannot come anywhere close to an answer
 
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That's the right appraoch, use the normal laws of motion - expect that you have an initial velocity.
The only force on the bomb is gravity vertically. So it has a constant vertical acceleration and a constant horizontal velocity.
 
So if the plane's velocity is V, the initial x component is, from magical trig, is V*sin(51)

the initial y component is V*cos(51)

you know time, it's t=4, and you know the horizontal distance is 570m

so -570=-1/2*g*t^2+V*cos(51)*t, remember it's negative 570 because the equation is normally d=1/2at^2+Vi*t, where d is final position - initial, and its final position is 0, initial is 570

You know everything in that equation except V, so solve for V, and that gives you part a)

using V you can find both Vx and Vy, and Vx let's you solve for b)

You still know t and now know both initial velocities and accelerations involved, so you can do c
 
The 'horizontal' distance is 570m, to get the vertical impact speed just use v = u + a t
 
i still cannot come up with the vertical component of velocity...
 
You need to find V first, then V*cos(51)=initial y velocity

The trig functions are reversed from your typical projectile motion because you're measuring from the vertical
 
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