Question

A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g=10 m/s².Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :
For last 6m distance travelled , s=6m
time=0.2s
initial velocity=?
a=g=9.8m/s2
s=ut+1/2at²
=6=u(0.2)+1/2 x 9.8 x0.2x0.2
6=0.2 u + 4.9 x 0.04
u=5.8/0.2
u=29m/s
For distance x, u=0m/s
v=29m/s [ because initial velocity will be final]
a=g=9.8m/s2
s= v²-u²/2a
=29²-0²/2x9.8
=29x29/19.6
s=42.5m
Total distance=42.5 +6=48.5 m

Answer 2

HEY!!

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☞Distance of the ball in 0.200 seconds = 6 m

☞Time, t = 0.200 s

☞Distance, s = 6 m

☞a = g = 10 m/s^2

☞Equation of motion = s = ut +(1/2) at^2

☞6= u ( 0.2 ) +1/2 × 10 × 0.04

☞u = 5.8/0.2=29 m/s

☞Let h be the height from which the ball is dropped.

☞u = 0 and v = 29 m/s

☞h=v2−u2/2a   
    
☞h=292−02/ 2×10 = 29×29/ 20=42.5 m

▶▶Total height = 42.5 + 6 = 48.5 m.