consider the motion in Y-direction
v₀ = initial velocity = 29 Sin62 = 25.6 m/s
a = acceleration = - 9.8 m/s²
t = time of travel
Y = vertical displacement = - 0.89 m
using the equation
Y = v₀ t + (0.5) a t²
- 0.89 = (25.6) t + (0.5) (- 9.8) t²
t = 5.3 sec
consider the motion along the horizontal direction :
v₀ = initial velocity = 29 Cos62 = 13.6 m/s
a = acceleration = 0 m/s²
t = time of travel = 5.3 sec
X = horizontal displacement =?
using the equation
X = v₀ t + (0.5) a t²
X = (13.6) (5.3) + (0.5) (0) t²
X = 72.1 m
d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m
t = time taken = 5.3 sec
v = speed of center fielder
using the equation
v = d/t
v = 34.9/5.3
v = 6.6 m/s
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x =
x = 
let's calculate
x =
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
I think the answer you're looking for is -are randomly oriented. if not sorry... i tried.
Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:
A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
ΔP=20 kg.m/s
Explanation:
Given data
Mass m=0.2 kg
Initial speed Vi=-44.5m/s
Final speed Vf=55.5 m/s
Required
Change in momentum ΔP
Solution
First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

Now we need to find the initial momentum
So

Substitute the given values

Now for final momentum

So the change in momentum is given as:
ΔP=P₂-P₁
![=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s](https://tex.z-dn.net/?f=%3D%5B%2811.1kg.m%2Fs%29-%28-8.9kg.m%2Fs%29%5D%5C%5C%3D20kg.m%2Fs)
ΔP=20 kg.m/s