What is the speed of an apple dropped from a tree after 1.5 second .? What distance will it covers during this time
Answers
Answered by
68
v = u + at
= 0 + (9.8 m/s² × 1.5 s)
= 14.7 m/s
Speed after 1.5 seconds is 14.7 m/s
S = ut + 0.5at²
= 0 + (0.5 × 9.8 m/s² × (1.5 s)²)
= 11.025 m
Distance travelled during this time is 11.025 m
= 0 + (9.8 m/s² × 1.5 s)
= 14.7 m/s
Speed after 1.5 seconds is 14.7 m/s
S = ut + 0.5at²
= 0 + (0.5 × 9.8 m/s² × (1.5 s)²)
= 11.025 m
Distance travelled during this time is 11.025 m
Answered by
40
Given ,
g=10m(approximate value of 9.8 ,constant value )
Initial velocity (u) = 0m/s,
Time (t) = 1.5sec,
S=?
V=?
We know that ,
V=u+gt
=0+10(1.5)
=12m/s
S=ut+1/2at( t square )
=0(t)+1/2gt( t square )
=1/2 (10)(1.5)1.5 (2 , 10 cancellation)
=11.25m
.
. . V=15m/s , S = 11.25m/s
Hope it helps you
g=10m(approximate value of 9.8 ,constant value )
Initial velocity (u) = 0m/s,
Time (t) = 1.5sec,
S=?
V=?
We know that ,
V=u+gt
=0+10(1.5)
=12m/s
S=ut+1/2at( t square )
=0(t)+1/2gt( t square )
=1/2 (10)(1.5)1.5 (2 , 10 cancellation)
=11.25m
.
. . V=15m/s , S = 11.25m/s
Hope it helps you
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