please solve question 4 it's urgent.
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(a) let the velocity be u.
if he catched the ball after 8 seconds so time taken for the ball to reach at its maximum point would be half of it i.e. 4 seconds.
using,v=u+at
v=0(at maximum height ofcourse)
0=u+(-g)(4)[-g because direction of displacement(upward) and direction of gravity(downward) is opposite]
u=4g=4*9.8=calculate it.
(b)at maximum height velocity of the ball would be zero.
(c)s=ut+(1/2)*at^2=(4g)(4)+(1/2)*(-g)(4*4)
=16g+(-8g)
=8g=8*9.8 answer.
if he catched the ball after 8 seconds so time taken for the ball to reach at its maximum point would be half of it i.e. 4 seconds.
using,v=u+at
v=0(at maximum height ofcourse)
0=u+(-g)(4)[-g because direction of displacement(upward) and direction of gravity(downward) is opposite]
u=4g=4*9.8=calculate it.
(b)at maximum height velocity of the ball would be zero.
(c)s=ut+(1/2)*at^2=(4g)(4)+(1/2)*(-g)(4*4)
=16g+(-8g)
=8g=8*9.8 answer.
Ruby04:
What it's maximum height ?
for maximum height ,use the formula:s=ut+(1/2)*at^2
mark me as brainliest.
Answer is not coming. please solve it
ok.
i have solved that in my qn.
answer*
thanku soo much
no problem:)feel free to ask anything.
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