Question

Height of a tower is 80 m. Two balls A and B are thrown simultaneously. Ball A is thrown upwards with speed u from top of tower while ball B is thrown upwards with speed of 50 m/s from the foot of tower

Ball A will reaches ground in 8 sec if ball do not collide then u is equal to -

(a) 20 m/s

(b) 25 m/s

(c) 30 m/s

(d) 35 m/s

Answer 1

Answer:

let the accelerations of both ball be aA=−g,aB=−g

and let the velocities be vA=20m/s,vb=−20m/s

relative acceleration=aA−aB=0

relative velocity=vA−vB=40m/s

s=ut+21at2

s=40m,u=40m/s,a=0

∴t=vs=4040=1s

For distance from ground we will observe particle A

s=ut+21at2

v=20m/s,t=1s,a=−g=−9.8m/s2

s=20−21×9.8×12=25m

Explanation:

HOPE IT HELPS YOU OUT ☺

HAVE A NICE DAY

Answer 2

Answer:

30 m/s

Explanation:

v = u + at

here , v = u + gt    -----(1)

now we know 3rd equation of motion , v² = u² + 2as

so here v² = u² + 2gs

(u+80)² = u²  -1600  (from (1) , g=10m/s² , s = 80m)

u² + 6400 + 160u = u² + 1600

160u = -4800

u = -30 m/s

|u| = 30 m/s

hence u = 30 m/s