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:

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

Answer 2

This is a physics excercise culled from the topic: Motion in a Plane, where the calculation of a projectile in motion is computed.

There are certain factors that have to be clarified.

1. The coordinate origin starts at ground level below the point of realease.
2. We can thus state that ∅₀ = (37.0⁰), where ∅ is the angle measured from +x.
3. Given that the angle Ч₀ = 53⁰ since the occurence is calibrated from -y direction.
4. It is also to be noted that the kinematics computations assume that vertical and horizontal directions are mutually exclusive and do not affect one another.

Solution is as follows

A) The projectiles initial speed is same as the plane's speed at the time of release. Given that:

Y₀ = 730m and Y = 0 at t = 5.00s. First, we need to find V₀.

Y-Y₀ = (V₀sin∅₀) t - $\frac{1}{2}$gt² ⇒ 0.730m = V₀sin (-37.0⁰) x (5.00s) - $\frac{1}{2}$(9.80m/s²)(5.00s)²

This translates to V₀ = 202 m/s.

(B) The distance travelled horizontally will be given as 806m.

Assuming distance travelled is G,
G = Vₓt = (V₀CosФ₀)t = ((202m/s)cos(-37.0⁰))(5.00s)

G = 806m

(C) Assume that the x components of the velocity (right before the impact) is given as Vx,

Vx = V₀cosФ₀

= ((202m/s) cos (-37.0⁰)

Vx= 161m/s.

(D) Where the Y component of the velocity just before impact is given as V$_{y}$,

V$_{y}$ = V₀SinФ₀ - gt

= (202m/s)Sin( - 37.0⁰) - (9.8m/s²)(5.00)

V$_{y}$= 171m/s.

For more excercises related to Projectile in Motion, please see the link below:

https://brainly.com/question/24888457