1. Two balls are shot with the same initial velocity - one is shot horizontally

Question

1. Two balls are shot with the same initial velocity - one is shot horizontally and the other is shot at an angle of 10 degrees above the horizontal. Which ball goes further? Explain your reasoning.

2. For a ball shot horizontally with a speed of v0 from a height of y0, what is the expression for the horizontal displacement of the ball in terms of v0, y0, and g (no time!)?

3. Use your expression from the previous question to find the horizontal displacement of a ball with an initial speed of 5 m/s and an initial height of 1.75 m being shot horizontally.

Explanation / Answer

1.

#Since the throwing of a ball at an angle is projectile motion it will have two components . One horizontal component and another vertical component. The horizontal component will decide the horizontal distance of the ball and the vertical component will provide the time to the ball to fall on the ground . In this case when the ball is shot horizontally so it's vertical component will be zero as compared to ball that is shot at 10 degree angle. So the horizontally shot ball will fall on the ground in less time than the another ball shot at 10 degree. Since the initial horizontal velocity is almost same for both, the ball shot at the 10 degree will cover more time.

For example- let 10 is the velocity. Then forhorizontally ball velocity is 10. For 10 degree horizontal velocity is 10cos10=9.848.

Now the time taken by the ball shot at 10 degree to go up and return to same height as that of initial position will be equal to t = 2×10sin10 / 9.8 = 0.3543 second.

So this ball will take 0.3543 more seconds to fall to ground. Let horizontally shot ball took 2 second .then distance travelled by that ball is = 10 ×2 = 20

So ball shot at 10 degree will take 2+0.3543= 2.3543 seconds

So distance travelled by it is = 2.3543× 9.848= 23.18

So this proves that ball shot at angle 10 degree will cover more distance.

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