Question

A projectile is launched at 37 degrees to the horizontal with a speed of 200 m/s from position A. How many seconds elapse after launch of the projectile before it returns to its original elevation at C?

a. 12 s
b. 16 s
c. 41 s
d. 33 s
e. 24 s

Answer 1

Answer:

e. 24 s

Explanation:

Time of flight of a projectile: This is the time taken for a projectile to return to the same plane at which it was projected. The time of flight of a projectile is expressed as,

T = 2Usinθ/g ........................ Equation 1

Where T = Time of flight, U = initial velocity, θ = angle of projection to the horizontal, g = acceleration due to gravity.

Given: U = 200 m/s, θ = 37°, g = 9.81 m/s²

Substitute into equation 1,

T = 2(200)(sin37°)/9.81

T = 400(0.60)/9.81

T = 24.46

T ≈ 24 s.

The right option is e. 24 s

Answer 2

The projectile time of the flight is 24.56 seconds.

Option E is the correct answer.

Projectile Time of Plane

The projectile time of the flight is the time from when the flight is projected to the time it reaches the surface. The formula to calculate the projectile time of the flight is given below.

$T = 2\times \dfrac {u\times sin \theta}{g}$

Where T is the projectile time, u is the initial speed of the flight, g is the gravitational acceleration and $\theta$ is the angle of projection.

Given that the angle of projection is 37 degrees and the initial speed is 200 m/s. Substituting the values in the above formula, we get the projectile time is,

$T = 2\times \dfrac {200\times sin 37^\circ} { 9.8}$

$T = 24.56 \;\rm s$

Hence we can conclude that the projectile time of the flight is 24.56 seconds. Option E is the correct answer.

To know more about the projectile time, follow the link given below.

https://brainly.com/question/11066165?.