A ball is thrown vertically upwards, moves according to the law `s=13.8t-4.9t^(2)` where s is in metres and t is in seconds.
(i) Find the acceleration at t = 1
(ii) Find velocity at t = 1
(iii) Find the maximum height reached by the ball?

A particle moves along a line according to the law s(t)=2t^(3)-9t^(2)+12t-4 , where tge0 . Find the particle's acceleration each time the velocity is zero.

A ball is thrown up at a speed of 4.0 m/s. Find the maximum height reached by the ball. Take g=10 m//s^2 .

A particle moves in circle of radius 1.0 cm at a speed given by v=2.0 t where v is in cm/s and t in seconeds. A. find the radia accelerationof the particle at t=1 s. b. Findthe tangential acceleration at t=1s. c.Find the magnitude of the aceleration at t=1s.

A ball is thrown vertically upwards with the speed of 19.6 ms^(-1) from the top of building and reaches the earth in 6 s. Find the height of the building .

The equation of motion is given by s=10t^(2)+60t . Where s is displacement is kilometre and t is time in hour find Velocity and acceleration at any point.

The equation of motion is given by s=10t^2+60t , where s is displacement in kilometre and t is time in hour. Find the velocity and acceleration at any point.

A particle moves along a line according to the law s(t)=2t^(3)-9t^(2)+12t-4 , where tge0 . At what times the particle changes direction?

A ball thrown vertically upwards with a speed of 19.6ms^(-1) from the top of a tower returns to the Earth in 6s. Find the height of the tower. (g=9.8m//s^(2))