- Suppose that a ball is dropped from the upper observation deck of a building 450m above the ground. If the equation of the speed is \(s(t)=4.9t^2\), what is the velocity of the ball after 5 seconds. How fas is the ball travelling when it hits the ground?
- The equation \(s(t) = −4.9 t 2 + 49 t + 15\) gives the height in meters of an object after it is thrown vertically upward from a point 15 meters above the ground at a velocity of 49 m/sec. How high above the ground will the object reach?
Solution

At highest height

v=0

s'=0

t=5 second

the heigh is s(5) - If a ball is thrown into the air with a velocity of 40 ft/s, its height after t seconds is given by \(y=40t-16t^2\), find the velocity when t=2
- The position of a particle is giveen by \(f(t)=\frac{1}{1+t}\). Find the velocity and speed after 2 seconds.

# Distance, Velocity and Acceleration

By
Bed Prasad Dhakal
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