A ball thrown upward satisfy the equation s=160 t – 20 t^2, where S is in m (meter) and t is in sec. what is the maximum height achieved by the ball ?
(a) 80
(b) 160
(c) 240
(d) 320
According to me..
Here s=height reached by the ball.
t=time taken by the ball to reach the particular height.
A/c to equation of motion;
s=ut+1/2at^2 &
v=u+at
Here, a=acceleration due to gravity, i.e., g=9.8m/s^2
Since the acceleration is is the opposite direction, we have to use the equations of motions as-
s=ut-1/2gt^2 &
v=u-gt
Here v=final velocity
Since at maximum height the stone will have a velocity of zero (since it will stop at the highest point).
So, v=0
than
can anyone tell me where I am wrong ?