As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]



Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Answer:
2.26 s
Explanation:
Let's take down to be positive.
Given (in the y direction):
Δy = 25 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
25 m = (0 m/s) t + ½ (9.8 m/s²) t²
25 = 4.9t²
t = 2.26 s
If the ball instead had an initial horizontal velocity of 5 m/s, its initial vertical velocity is still 0 m/s. So the time to fall is still 2.26 s.
The period T of a pendulum is given by:

where L is the length of the pendulum while

is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is

. Using this data, we can solve the previous formula to find L:
It would be funny because . I will not be good
Answer:
that one i know only pe not that sorry again