Question

A plane, diving with constant speed at an angle of 53.0° with the vertical, releases a projectile at an altitude of730 m. The projectile hits the ground 5.00 s after release. (a) What is the speed of the plane? (b) How far doesthe projectile travel horizontally during its flight? What are the (c) horizontal and (d) vertical components of itsvelocity just before striking the ground?

Answer 1

Answer:

A) V=198.75m/s

B) 793.7mts

C) Vx=158.72m/s

D) Vy=-168.71m/s

Explanation:

In order to answer the first question we need to calculate the initial speed of the projectile because it's the same speed of the plane.

$Vfy^2=Voy^2+2*a*d\\Vfy=0\\Voy=\sqrt{-2*(-9.8)*730} \\Voy=119.61m/s$

The velocity has to be negative because is going down so, Voy=-119.61m/s

now that we have the Y component of velocity we can get the plane velocity by:

the angle with the horizontal is ∝=-90+53=-37

$Voy=V*cos(\alpha )\\V=\frac{Voy}{sin(\-53})\\V=198.75m/s$

Now that we have the Initial velocity we can calculate the horizontal displacement:

$d=V*cos(\alpha )*t\\d=198.75*cos(-37)*5\\d=793.7mts$

Because the horizontal velocity remains the same, it is given by:

$Vx=V*cos(-37)=158.72m/s$

In order the calculate the Y componet we will use the next formula:

$Vfy=Vo+a*t\\Vfy=-119.71+(-9.8)*5\\Vfy=-168.71m/s$