To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to
![v_x = \frac{x}{t}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
![v_x = \frac{67}{4.5}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7B67%7D%7B4.5%7D)
![v_x = 14.89m/s](https://tex.z-dn.net/?f=v_x%20%3D%2014.89m%2Fs)
The vertical component of velocity is
![-h = v_y t -\frac{1}{2} gt^2](https://tex.z-dn.net/?f=-h%20%3D%20v_y%20t%20-%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
Here,
h = Height
g = Gravitational acceleration
t = Time
= Vertical component of velocity
![-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2](https://tex.z-dn.net/?f=-1.23%20%3D%20v_y%284.5%29-%5Cfrac%7B1%7D%7B2%7D%20%289.8%29%284.5%29%5E2)
![-1.23= 4.5v_y - 99.225](https://tex.z-dn.net/?f=-1.23%3D%204.5v_y%20-%2099.225)
![v_y = 21.77m/s](https://tex.z-dn.net/?f=v_y%20%3D%2021.77m%2Fs)
The direction of the velocity will be given by the tangent of the components, then
![tan\theta = \frac{v_y}{v_x}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_x%7D)
![\theta = tan^{-1} (\frac{21.77}{14.89})](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B21.77%7D%7B14.89%7D%29)
![\theta = 55.59\°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2055.59%5C%C2%B0)
The magnitude is given vectorially as,
![|V| = \sqrt{v_x^2+v_y^2}](https://tex.z-dn.net/?f=%7CV%7C%20%3D%20%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D)
![|V| = \sqrt{14.89^2 +21.77^2}](https://tex.z-dn.net/?f=%7CV%7C%20%3D%20%5Csqrt%7B14.89%5E2%20%2B21.77%5E2%7D)
![|V| = 26.37m/s](https://tex.z-dn.net/?f=%7CV%7C%20%3D%2026.37m%2Fs)
Therefore the angle is 55.59° and the velocity is 26.37m/s
C.) <span>The total mass of an object can be assumed to be focused at one point, which is called its center of "Mass"
Hope this helps!</span>
Answer:
It changes into a completely different element
The answer is centimeters.
Answer:
The two balls meet in 1.47 sec.
Explanation:
Given that,
Height = 25 m
Initial velocity of ball= 0
Initial velocity of another ball = 17 m/s
We need to calculate the ball
Using equation of motion
![s=ut+\dfrac{1}{2}gt^2+h](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%2Bh)
Where, u = initial velocity
h = height
g = acceleration due to gravity
Put the value in the equation
For first ball
....(I)
For second ball
....(II)
From equation (I) and (II)
![-\dfrac{1}{2}gt^2+25=17t-\dfrac{1}{2}gt^2+0](https://tex.z-dn.net/?f=-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%2B25%3D17t-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%2B0)
![t=\dfrac{25}{17}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B25%7D%7B17%7D)
![t=1.47\ sec](https://tex.z-dn.net/?f=t%3D1.47%5C%20sec)
Hence, The two balls meet in 1.47 sec.