Solution:
According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.
So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.
v_f=v_o + at ……..(a)
[where v_f and v_o are final velocity and initial velocity, respectively]
Now ,
Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.
Applying this value in equation (a)
v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction
For calculating the magnitude of the equation we have to square root the given value
(16.6i - 17.165y)
\\
\left | V \right |=sqrt{16.6^{2}+17.165^{2}}\\ =
\sqrt{275.56+294.637225}\\=
\sqrt{570.197225}\\=
23.87[/tex]
Answer: 0m/s²
Explanation:
Since the forces acting along the plane are frictional force(Ff) and moving force(Fm), we will take the sum of the forces along the plane
According newton's law of motion
Summation of forces along the plane = mass × acceleration
Frictional force is always acting upwards the plane since the body will always tends to slide downwards on an inclined plane and the moving acts down the plane
Ff = nR where
n is coefficient of friction = tan(theta)
R is normal reaction = Wcos(theta)
Fm = Wsin(theta)
Substituting in the formula of newton's first law we have;
Fm-Ff = ma
Wsin(theta) - nR = ma
Wsin(theta) - n(Wcos(theta)) = ma... 1
Given
W = 562N, theta = 30°, n = tan30°, m = 56.2kg
Substituting in eqn 1,
562sin30° - tan30°(562cos30°) = 56.2a
281 - 281 = 56.2a
0 = 56.2a
a = 0m/s²
This shows that the trunk is not accelerating
The third one sliding friction
Explanation:
Time it takes the projectile to hit the ground after being thrown up:
√h/1/2a
√8/(.5)(9.81)
√8/4.905
√1.630988787
= 1.277101714
= 1. 28
hope this helps :)
