Question

Hey ..... plzzz solve ìt fast clearly and in a copy.

Answer 1

Hey sistah !!!!!

Here's your answer =>=>

Let's find ur answer with the help of equations of motion ===

There are 3 equations of motion

a) v = u + at

b) s = ut + 1/2 at^2

c) v^2 = u^2 + 2as

So first we will find the tym with the help of a) equation.

Let's start !_!

Given = v = 0m/s

u = 5m/s

a = -10m/s^2

________________________________________

v = u + at

0 = 5 - 10 × t

-5 = -10t

-5/-10 = t

1/2 = t

0.5 seconds = t

________________________________________

Now let's find the height/distance ===

You can find it either by b) equation or by c) equation.

First we will find it by c) equation ===

${v}^{2} = {u}^{2} + 2as \\ \\ {0}^{2} = {5}^{2} + 2 \times -10 \times s \\ \\ 0 = 25 - 20s \\ \\ -25 = -20s \\ \\ \frac{25}{20} = s \\ \\ 1.25 \: metre = s$

Now let's do it by equation b)

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ s = 5 \times \frac{1}{2} + \frac{1}{2} \times -10 \times { \frac{1}{2} }^{2} \\ \\ s = \frac{5}{2 } - \frac {5}{4} \\ \\ so by taking l.c.m. It can be solved. s = \frac{5}{4} \\ \\ s = 1.25 \: metre$

Hope it will help you ☆▪☆

Thanks ^_^

# Regards #

@Beautiful5225

Answer 2

Heya !!!

Here is Your answer.

Given :
A ball is thrown vertically upwards with velocity of 5 m/s.
So, u = 5 m/s.
The Ball teaches to height and it will surely stop there and then will return back.
So, v = 0
Also, Acceleration is 10 m/s in downward direction but the ball is going upwards. So, the acceleration will be negative.
a = -10 m/s².

Using First Equation of motion...
v = u + at
0 = 5 - 10t
-5 = -10t
t = 5/10 = 0.5 s.

Now, Using Second or Third Equation of Motion .....(we can use anyone of them)
s = ut + 1/2× at²
$s = 5 \times 0.5 + \frac{1}{2} \times -10 \times {(0.5)}^{2} \\ = 2.5 - 5 \times 0.25 \\ = 2.5 - 1.25 = 1.25\: m$

Hence, the highest point it will reach is 1.25 m and will take 0.5 s.

Hope You Got It