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Physics: Post your doubts here!

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Yea i know But WHYYY
We've got everything except displacement why can't we use this eq :(
See in such questions related to projectile motion, we need to deal with the horizantal and vertical components separately.
The ball is hit HORIZANTALLY, with a speed of 32 m/s. The HORIZANTAL distance is 5.5 m.


Since the ball is hit horizantally there is no VERTICAL omponent of speed, The acceleration is 10m/s. Acceleration due to gravity is itself a VERTICAL component and not a horizantal. We need to find the distance the ball has dropped. This is the vertical component of distance.

You cant use the equation you stated because:
if we were to use the equation for the horizantal component, we wouldnt be able to get an answer as the "a" in the formula will have no value. LIKE I SAID BEFORE ACCELERATION DUE TO GRAVITY IS PART OF THE VERTICAL COMPONENT ONLY.
 
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sum4.gif
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dp1.gif


Line 1: The change in momentum equals the mass times the change in velocity.

Line2: The change in velocity is the final velocity minus the original velocity.

Line 3: Distribute the m.

Line 4: The change in momentum equals the final momentum minus the original momentum. Not initial minus final.
 
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On a particular railway, a train driver applies the brake of the train at a yellow signal, a distance of
1.0km from a red signal, where it stops.
The maximum deceleration of the train is 0.2ms–2
.
Assuming uniform deceleration, what is the maximum safe speed of the train at the yellow signal?
A 20ms–1 B 40ms–1 C 200ms–1 D 400ms–1
 
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On a particular railway, a train driver applies the brake of the train at a yellow signal, a distance of
1.0km from a red signal, where it stops.
The maximum deceleration of the train is 0.2ms–2
.
Assuming uniform deceleration, what is the maximum safe speed of the train at the yellow signal?
A 20ms–1 B 40ms–1 C 200ms–1 D 400ms–1
A is a guess.
 
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