Question

A ball is thrown upward with a velocity u such that it travelled a height more than the height of a
building h. If t1, and t2 are the two times in which the displacement of the ball was h. Find
a) t1 + t2
b) t1*t2​

Answer 1

Topic :- Motion in staright line

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

A ball is thrown vertically upwards with Initial Velocity "u" and it is at certain height "h" (h < hₘₐₓ) for two times t₁ and t₂. [Acceleration due to gravity = -g ]

$\dag\:\underline{\tt By \: using \: second \: kinematical \: equation \: of \: motion \: we \: get : }$

$\longrightarrow\:\:\sf H = ut + \dfrac{1}{2} (-g)t^2$

$\longrightarrow\:\:\sf H = ut + \dfrac{( - g)}{2} t^2$

$\longrightarrow\:\:\sf H = ut - \dfrac{ g}{2} t^2$

$\longrightarrow\:\:\sf \dfrac{ g}{2} t^2 - ut + H = 0$

Comparing the above equation with quadratic equation we have :

$\longrightarrow\:\:\sf {ax}^{2} + bx + c$

Where,

• a = g/2
• b = -u
• c = H

$\dag\:\underline{\tt Sum \: of \: zeros : }$

$\longrightarrow\:\:\sf t_1 + t_2 = \dfrac{-b}{a}$

$\longrightarrow\:\:\sf t_1 + t_2 = \dfrac{ - (-u)}{ \dfrac{g}{2} }$

$\longrightarrow\:\:\sf t_1 + t_2 = \dfrac{ u}{ \dfrac{g}{2} }$

$\longrightarrow\:\: \underline{ \boxed{\sf t_1 + t_2 = \dfrac{ 2u}{g}}}$

$\dag\:\underline{\tt Product \: of \: zeros : }$

$\longrightarrow\:\:\sf t_1.t_2 = \dfrac{c}{a}$

$\longrightarrow\:\:\sf t_1.t_2 = \dfrac{H}{ \dfrac{g}{2} }$

$\longrightarrow\:\: \underline{ \boxed{\sf t_1.t_2 = \dfrac{2H}{ g }}}$