High School

We appreciate your visit to A projectile is fired with an initial speed of 37 6 m s at an angle of 43 6 above the horizontal on a long. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A projectile is fired with an initial speed of 37.6 m/s at an angle of 43.6° above the horizontal on a long, flat firing range.

Answer :

Final answer:

The maximum height reached by the projectile is 24.7 meters. The total time in the air is 5.38 seconds. The total horizontal distance covered (range) is 198.9 meters.

Explanation:

The maximum height reached by the projectile can be determined using the equations of projectile motion. The equation to calculate maximum height is given by:

h_max = (v^2*(sin^2(a)))/(2*g)

Using the given values of initial speed (v = 43.6 m/s) and launch angle (a = 43.2°), and taking the acceleration due to gravity (g) as 9.8 m/s^2, the maximum height can be found:

h_max = (43.6^2*(sin^2(43.2°)))/(2*9.8) = 24.7 m

The maximum height reached by the projectile is 24.7 meters.

The total time in the air can be determined using the equation:

t = 2*v*sin(a)/g

Substituting the given values, we get:

t = 2*43.6*sin(43.2°)/9.8 = 5.38 s

The total time in the air is 5.38 seconds.

The total horizontal distance covered, also known as the range, can be calculated using the equation:

R = v*cos(a)*t

Substituting the given values, we get:

R = 43.6*cos(43.2°)*5.38 = 198.9 m

The total horizontal distance covered (range) is 198.9 meters.

To determine the speed of the projectile 1.00 second after firing, we need to calculate the horizontal and vertical components of the velocity at that time. The horizontal component is given by:

Vx = v*cos(a)

Substituting the given values, we get:

Vx = 43.6*cos(43.2°) = 31.2 m/s

The vertical component is given by:

Vy = v*sin(a) - g*t

Substituting the given values, we get:

Vy = 43.6*sin(43.2°) - 9.8*1.00 = 14.6 m/s

The speed of the projectile 1.00 second after firing is:

V = sqrt(Vx^2 + Vy^2) = sqrt((31.2)^2 + (14.6)^2) = 34.2 m/s

The speed of the projectile 1.00 second after firing is 34.2 m/s.

The direction of the motion of the projectile 1.00 second after firing can be determined using the angle it makes with the horizontal. The angle is given by:

θ = tan^(-1)(Vy/Vx)

Substituting the given values, we get:

θ = tan^(-1)(14.6/31.2) = 24.1°

The direction of the motion of the projectile 1.00 second after firing is 24.1 degrees above the horizontal.

Learn more about Projectile Motion here:

https://brainly.com/question/29545516

#SPJ11

content loaded

A projectile is fired with an initial speed of 43.6 m/s at an angle of 43.2 ∘ above the horizontal on a long flat firing range.

Part A

Determine the maximum height reached by the projectile.

Part B

Determine the total time in the air.

Part C

Determine the total horizontal distance covered (that is, the range).

Part D

Determine the speed of the projectile 1.00 s after firing.

Part E

Determine the direction of the motion of the projectile 1.00 s after firing

Thanks for taking the time to read A projectile is fired with an initial speed of 37 6 m s at an angle of 43 6 above the horizontal on a long. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada