Question

A ball projected from ground vertically upward is at same height at time t1and t2.the speed of projection of ball is

Answer 1

Final Answer: $u= g (\frac{t_{1}+t_{2}}{2} )$ 
Assuming $t_{1} \neq t_{2}$ .
Steps:
1) Let the initial / projection speed be 'u'.
acceleration due to gravity, a = -g , 
Then,
As Displacement in y-direction is same .

According to question ,
 [tex]S_{y}=S_{y}' \\ \\ =\ \textgreater \ ut_{1} - \frac{1}{2}gt_{1}^{2} = ut_{2}- \frac{1}{2} g t_{2}^{2} \\ \\ =\ \textgreater \ u(t_{1}-t_{2}) - \frac{1}{2}g(t_{1}^{2}-t_{2}^{2}) = 0 \\ \\ =\ \textgreater \ u - \frac{1}{2}g(t_{1}+t_{2})=0 \:\:\: (since \:\:\:t_{1} \neq t_{2})\\ \\ =\ \textgreater \ u = \frac{1}{2}g(t_{1}+t_{2}) [/tex]

Hence , Projection speed of ball is 

$\boxed{\frac{1}{2}g(t_{1}+t_{2})}$