Answer: Highest point from the ground that ball reaches = 5.102
Explanation:
Formula -
v² - u² = 2as
final velocity v = 0
initial velocity u = 10m/s
acceleration a = g (gravity)=9.8
distance s = 3m+x (x is the height after 3m that object reaches)
0-(10)² = 2 (-g)*(3+x)
-100 = -2g*(3=x)
(3+x)=100/2g
3+x = 100/2*9.8 = 100/19.6 = 5.102
x=5.102-3
x=2.102
So, the highest point will be
3+x = 3+2.102
=5.102m
Answer:
<em>The range of the ball is 11.6 meters</em>
Explanation:
<u>Projectile Motion
</u>
It's the type of motion that experiences an object launched with an initial angle and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and g the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
The football is kicked with an initial speed of vo=11 m/s at an angle of θ=35°.
Calculating the range:
d = 11.6 m
The range of the ball is 11.6 meters
His answer looks correct, the total N acting on this box is 2 N. from the applied force.
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃
We divide by (k * q3) on both sides of the equation
q₁ = + 1.25 nC