Two particles Q and P are shot up vertically.P was first shot with velocity of40m/s and after4s.Q was also shot up.If upon meeting the velocity ofP is 15m/s.determine(1)where they met.(2)the velocity with which Q was shot up.Take g= 10m/s2
PARTICLE P:
Tr = (Vf-Vo)/g,
Tr = (0-40) / -10 = 4.0 s. = Rise time.
hmax = Vo*t + 0.5g*t^2,
hmax = 40*4 - 5*4^2 = 80 m.
P is falling when Q is shot up:
1. h = hmax - (Vf^2-Vo^2)/2g,
h = 80 - ((15)^2-0) / 20 = 68.75 m.
Tf = (Vf-Vo)/g = (15-0) / 10 = 1.5 s. =
Fall time. = Rise time(Tr) for Q.
PARTICLE Q:
2. h = Vo*t + 0.5g*t^2 = 80-68.75 =11.25
Vo*1.5 - 5*(1.5)^2 = 11.25,
Vo*1.5 - 11.25 = 11.25,
Vo*1.5 = 22.5,
Vo = 22.5 / 1.5 = 15 m/s.
Excellent
I want to study with you to better my grades. I want to be a surgeon
For more improvement on my education
To determine where particles Q and P met, we need to first calculate the time it took for both particles to meet. Since particle Q was shot after particle P, we can assume that Q took t seconds to reach the point where the two particles met.
Let's solve the problem step by step:
Step 1: Determine the time it took for particle P to reach the meeting point.
Given:
Velocity of P, vP = 15 m/s
Acceleration due to gravity, g = 10 m/s^2
Using the kinematic equation:
vf = vi + at
where,
vf = final velocity
vi = initial velocity
a = acceleration
t = time
When particle Q meets particle P, both particles will have the same final velocity. Therefore, we can equate their final velocities:
vf(P) = vf(Q)
15 m/s = 40 m/s - 10 m/s^2 * tP
Solving for tP:
-10tP = -25
tP = 2.5 s
So, it took 2.5 seconds for particle P to reach the meeting point.
Step 2: Determine the initial velocity of particle Q.
Using the kinematic equation:
vf = vi + at
When particle Q meets particle P, both particles will have the same final velocity. Therefore, we can equate their final velocities:
vf(P) = vf(Q)
15 m/s = vi(Q) + 10 m/s^2 * tQ
Since particle Q was shot after particle P, we can calculate the time it took for Q to reach the meeting point by subtracting tP from tQ:
tQ = tP + 4s
tQ = 2.5s + 4s
tQ = 6.5s
Substituting the values back into the equation:
15 m/s = vi(Q) + 10 m/s^2 * 6.5s
Solving for vi(Q):
vi(Q) = 15 m/s - 65 m/s
vi(Q) = -50 m/s
The negative sign indicates that particle Q was shot downward. Therefore, the initial velocity of particle Q was -50 m/s.
To summarize:
(1) The particles met after 2.5 seconds at the same height.
(2) Particle Q was shot downward with an initial velocity of -50 m/s.