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Question

Two balls are thrown from the top of a tower at a speed $$40\ m/s$$. One is thrown $$30^{\circ}$$ above horizontal and one $$30^{\circ}$$ below horizontal. Find the time interval between the instants the balls strike the ground in seconds $$(g = 10\ m/s^{2})$$.


A
2 sec
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B
4 sec
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C
5 sec
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D
6 sec
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Solution

The correct option is B $$4\ sec$$
Let vertical velocity of ball which is thrown above horizontal $$v_A= 40sin(30^o)=20ms^{-1}$$, 
Let vertical velocity of ball which is thrown below horizontal $$v_B= 40sin(30^o)=20ms^{-1}$$, 
We know in the projectile when ball comes down after reaching the highest point its magnitude of velocity is same as when it goes upward at same point. 
$$\therefore $$ The velocity of ball which is thrown upward is same as velocity of ball which is thrown downward at top of tower when ball is returned at top of tower. So from there both ball takes same time so the time difference is the time taken by ball during upward journey to the top of tower.

At maximum height, Velocity $$v=0$$
Using $$v=u+at$$  $$\Rightarrow 0=20-gt$$  $$\Rightarrow t=2 s$$
$$\Rightarrow $$ Total time for upward journey$$= 2t= 4s$$, 
$$\therefore $$ Time difference$$=4s$$.

Physics

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