a ballon with a basket with a man inside starts from rest and rises vertically with a uniform acceleration of 4.9m/s2 the man releases a ball 2 second after the ballon leaves the ground the maximum height above the ground attained by the ball is???

Question

experimentx · Answer

calculate final velocity, v = u + at , u=0, t=2, a=4.9

and use the prevous v as u here, v^2 = u^2 + 2as, v=0, a = -g, find s

anonymous · Answer

plz can u solve it

anonymous · Answer

option are   (a) 9.8    (b)   14.7       (c)  19.6      (d)  24.5 and the answer is b can u solve it

experimentx · Answer

v= 9.8 => u  = 9.8

0 = 9.8^2 - 2* 9.8 * s => s = 9.8/2 = 4.9 m


Looks like we also have to consider the distance traveled at 2s before
s= ut + 1/2 at^2 = 0 + 1/2 * 4.9*2^2 = 9.8 m

add those you get, 4.9 + 9.8 and get your answer.

anonymous · Answer

s=ut+1/2at$$t ^$$
s=0x2+1/2x10x4
s=0+20
s=20m