Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe
2 answers:
Answer:
(1) 14.12 m/s
Explanation:
Given:
= initial speed of the ball = 16 m/s
= angle of the initial speed with the horizontal axis =
= initial height of the ball from where Julie throws the ball = 1.5 m
= final position of the ball where Sarah catches the ball = 1.5 m
Let us assume the following:
= horizontal component of the initial speed
= vertical component of the initial speed
= horizontal acceleration of the ball
= vertical acceleration of the ball
The given problem is projectile motion. When the ball is thrown from the air with a speed of 16 m/s at an angle 28 degree with the horizontal axis. When the ball is in the air, it experiences an only gravitational force in the downward direction if we ignore air resistance on the ball.
This means if we break the motion of the ball along two axes and study it, we have a uniform acceleration motion in the vertical direction and a zero acceleration motion along the horizontal.
Since the ball has a zero acceleration motion along the horizontal axis, the ball must have a constant speed along the horizontal at all instant of time.
Let us find out the initial velocity horizontal component of the velocity of the ball. which is given by:
As this horizontal velocity remains constant in the horizontal motion at all instants of time. So, the horizontal component of the ball's velocity when Sarah catches the ball is 14.12 m/s.
Hence, the horizontal component of the ball's velocity when the ball is caught by Sarah is 14.12 m/s.
You might be interested in
Answer:
See explanation
Explanation:
Given:-
- The DC power supply, Vo = 600 V
- The resistor, R = 1845 MΩ
- The plate area, A = 58.3 cm^2
- Left plate , ground, V = 0
- The right plate, positive potential.
- The distance between the two plates, D = 0.3 m
- The mass of the charge, m = 0.4 g
- The charge, q = 3*10^-5 C
- The point C = ( 0.25 , 12 )
- The point A = ( 0.05 , 12 )
Find:-
What is the speed, v, of that charge when it reaches point A(0.05,12)?
How long would it take the charge to reach point A?
Solution:-
- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.
E = Vo / D
E = 600 / 0.3
E = 2,000 V / m
- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:
Fe = E*q
Fe = (2,000 V / m) * ( 3*10^-5 C)
Fe = 0.06 N
- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:
Fnet = m*a
Where, a : The acceleration of the object/particle.
- The only unbalanced force acting on the particle is (Fe):
Fe = m*a
a = Fe / m
a = 0.06 / 0.0004
a = 150 m/s^2
- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:
s = C - A
s = ( 0.25 , 12 ) - ( 0.05 , 12 )
s = 0.2 m
- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:
vf^2 = vi^2 + 2*a*s
vf^2 = 0 + 2*0.2*150
vf = √60
vf = 7.746 m/s
- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:
vf - vi = a*t
t = ( 7.746 - 0 ) / 150
t = 0.0516 s
- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:
- Where, The constant c : The capacitance of the capacitor.
- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:
- Again, apply the Newton's second law of motion and determine the acceleration (a):
Fe = m*a
a = Fe / m
- Where the acceleration is rate of change of velocity "dv/dt":
- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:
Where, eo = 8.854 * 10^-12 .... permittivity of free space.
- The differential equation turns out ot be:
- Separate the variables the integrate over the interval :
t : ( 0 , t )
v : ( 0 , vf )
Therefore,
- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:
- The final velocity of particle while charging the capacitor would be:
vf = 7.264 m/s ... slightly less for the fully charged capacitor