Question

A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal. The projectile lands on the ground 5.0 seconds later. Neglecting air resistance, calculate the projectile's range and draw a projectile path.

Answer 1

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

$v_x = v_i cos \theta$

where

$v_i = 40 m/s$ is the initial velocity

$\theta=20^{\circ}$ is the angle of projection

Substituting,

$v_x = (40)(cos 20^{\circ})=37.6 m/s$

And therefore, the range of the projectile is:

$d=v_x t = (37.6)(5.0)=188 m$

Learn more about projectile motion:

brainly.com/question/8751410

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