Question

Someone throws their laptop with a horizontal component of velocity of 25 m/s. It takes the laptop 3.00 seconds to come back to the original height. Calculate the horizontal range and the initial vertical component of velocity

Answer 1

1) The horizontal range of the laptop is 75.0 m

2) The initial vertical component of the velocity is 14.7 m/s

Explanation:

1)

The motion of the laptopin 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  

To find the horizontal range of the laptop, we just need to analyze its horizontal motion.

We know that:

- The laptop moves horizontally with a constant velocity of $v_x = 25 m/s$

- The time of flight of the laptop is $t=3.00 s$

So, the horizontal range of the laptop is given by

$d=v_x t = (25)(3.00)=75.0 m$

2)

The total time of flight of a projectile is given by

$t=\frac{2u_y}{g}$

where

$u_y$ is the initial vertical component of hte velocity

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

Here we know that the time of flight is

$t=3.00 s$

Therefore, we can solve the formula for $u_y$, to find the initial vertical velocity:

$u_y = \frac{tg}{2}=\frac{(3.00)(9.8)}{2}=14.7 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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