- Let, the maximum height covered by projectile be


- Projectile is thrown with a velocity = v
- Angle of projection = θ
- Velocity of projectile at a height half of the maximum height covered be

______________________________
Then –










- Now, the vertical component of velocity of projectile at the height half of
will be –


Therefore, the vertical component of velocity of projectile at this height will be–
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Answer:
How much electricity the appliance can hold, the number of hours the appliance is used in a day, and how many days it is used of the year.
Explanation:
Once we find all these things its simple math to figure out how many watts the appliance uses.
Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0

t = 1.4 second