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3 years ago

Projectile SHOW WORK WILL MARK BRANLIEST (Draw Picture and Label)

Physics

1 answer:

m_a_m_a [10]3 years ago

8 0

a) The horizontal distance covered by the projectile is 600 m

b) The projectile reaches its maximum height after 3.00 s

c) The altitude of the highest point is 44.1 m

Explanation:

The motion of the projectile consists of two independent motions:

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

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this part A, we just need to analyze the horizontal motion. We know that:

The projectile travels horizontally with a constant velocity ov $v_x = 100 m/s$
The total time of flight of the projectile is $t=6.00 s$

Therefore, the horizontal distance covered by the projectile is given by

$x=v_x t$

And substituting, we find

$x=(100)(6.0) = 600 m$

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

$s=u_y t + \frac{1}{2}at^2$

where:

$u_y$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

t is the time

s is the vertical displacement

We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for $u_y$ , we get

$u_y = - \frac{1}{2}at=-\frac{1}{2}(-9.8)(6)=29.4 m/s$

The vertical velocity then as a function of t is given by

$v_y = u_y + at$

And at the maximum height, it becomes zero: $v_y = 0$ . Substituting and solving for t, we find the time at which the projectile reaches the maximum height:

$t=-\frac{u_y}{a}=-\frac{29.4}{-9.8}=3.00 s$

To find the altitude of the highest point in the path, we use again the equation:

$s=u_y t + \frac{1}{2}at^2$

where

$u_y = 29.4 m/s$ is the initial vertical velocity

t = 3.00 s is the time at which the projectile reaches the highest point

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Substituting the values, we find

$s=(29.4)(3.00)+\frac{1}{2}(-9.8)(3.00)^2=44.1 m$

So, the highest point is at 44.1 m above the ground.

Learn more about projectiles:

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