Explanation:
It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m
Using the theorem of kinetic energy
1/2mVf² - 1/2mVi²= WF + Wp, Wp=0
WF = F. AB, AB=5m and F= 40N, m=20kg
so the final kinetic is KEf= 1/2mVf² = WF =<span>F. AB= 40*5=200J
</span>
the final velocity is 1/2mVf² <span>=200, implies Vf= sqrt(20)=2sqrt(5)m/s</span>
Answer:
You need to have a good stance, release, power, speed, body, movement, and aim. You should be able to throw an effective pass or throw by using your body and having the correct amount of balance and form.
Explanation:
