A projectile launched from the ground with an initial velocity 20sqrt5 m/s , just clears two walls each 50m high and seperated by 100m what is the time of passage b/w the walls?

Question

mathslover · Answer

first write given thingies : 
u = 20sqrt5 m/s
vertical height = 50 m 
horizontal height ( between walls) =100m

mathslover · Answer

s = ut+1/2 at2 ..

mathslover · Answer

hmn wait

mathslover · Answer

i got it --> i think

mathslover · Answer

attachment

anonymous · Answer

@mathslover  it is the prob frm motion in 2d  so resolving vectors is imp

mathslover · Answer

like using eqn of motion ..

mathslover · Answer

ok first calculate time of flight

mathslover · Answer

...... confused........

dls · Answer

Can you assume the walls to be the highest point of the body and apply the formula of max height?

anonymous · Answer

it is said it hust cleared so i think we can take it as Hmax

dls · Answer

Apply the formula and tell me the answer.

anonymous · Answer

@DLS  we need to find what is the time of passage b/w the walls?

maheshmeghwal9 · Answer

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Let the two pillars be covered in t1 & t2 seconds.


then we get 3 equations

$$50=20\sqrt{5}\sin \theta t_1-\frac{1}{2} g t_1^2.$$............ (1.) eq.
$$50 = 20 \sqrt{5} \sin \theta t_2-\frac{1}{2} g t_2^2.$$.............(2) eq.

$$\color{red}{100=20 \sqrt{5}\cos \theta (t_2-t_1).}$$..............(3.) eq.