Question

A ball is thrown vertically upward from the top of a tower 40m high with velocity 10m/sec find time when it strikes ground

Answer 1

GIVEN:-

• $\rm{Initial\:Velocity = +10m/s}$
• $\rm{Acceleration = -10m/s^{-2}}$

• $\rm{height\:of\:tower = 40m}$

TO FIND:-

• Time when it strikes ground

FORMULAE USED:-

• ${\boxed{\rm{S = ut+ \dfrac{1}{2}\times{a}\times{(t)^2}}}}$

Now,

$\implies\rm{ S = ut+ \dfrac{1}{2}\times{a}\times{(t)^2}}$

$\implies\rm{ 40 = 10\times{t} - \dfrac{1}{\cancel{2}}\times{\cancel{10}}\times{(t)^2}}$

$\implies\rm{ -40 = 10t - 5t^2}$

$\implies\rm{ -8 = 2t - t^2}$

$\implies\rm{ t^2 - 2t - 8= 0}$

$\implies\rm{ t^2 -4t + 2t - 8 = 0}$

$\implies\rm{ t( t -4) + 2(t-4) = 0}$

$\implies\rm{ (t-4) = 0 = 4s}$.

Hence, The time by it is 4s.