Q&A for How to Solve a Projectile Motion Problem

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What is uniform motion?

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The motion in which the velocity is constant at each and every interval as there is no acceleration in the x-axis component.

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In a projectile motion, the first object is thrown vertically upward with an initial velocity of 40 m/s. After two seconds, a second object is thrown upward with the same velocity. What is the height were they meet?

Cabbache

Community Answer

The height would be 76.644 meters. Draw a graph of both objects on the same axis showing the height of object against time. You will see 2 parabolas that intersect at one point. You are asked to find the height of that point. You can do so if you equate the equations of each parabola. The equations would be based on s = ut + (at^2)/2 where s is the height, u is the initial velocity, t is the time elapsed and a is the acceleration due to gravity. The first parabola would be s = 40t + (9.81t^2)/2, and the second parabola would be s = 40(t - 2) + (9.81(t - 2)^2)/2.

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How you can change km/h to m/s?

Startrekker

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To convert km/h to m/s, multiply it by 5/18. For example, 36 km/h = 36*5/18 m/s = 10m/s. To convert m/s to km/h, multiply it by 18/5 (just the opposite of the above).

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What are the two independent problems in the motion of a projectile?

Startrekker

Community Answer

The motion in the x-axis is independent of motion in the y-axis. Calculate them independently. for example, the time of flight depends only on the vertical component of the initial velocity. No matter how large is the horizontal velocity, the object will land at the same time. (Although since the earth is round, the time of flight increases. We say the time of flight is same only for small distances).

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A projectile is launched vertically at 100m/s. If air resistance can be neglected, at what speed does it return to its initial level?

Startrekker

Community Answer

Remember that if there is a uniform acceleration of the object, and the initial velocity is opposite to acceleration, the object will return after some time at the height of the initial point with the same velocity at which it was thrown. Initial velocity=u, Acceleration = -g, displacement = 0, v=? v^2 = u^2 + 2gs = u^2 + 0 (since s = 0) v^2 = u^2 IvI = IuI. It will return to the initial level at the same "speed".

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How do I know whether to make velocity negative or positive when assigning directions?

Startrekker

Community Answer

Decide a coordinate system. It will be 2 mutually perpendicular lines. Place the origin as per convenience. If the object is going in +x, assign x component of velocity as +ve. If -x, then x component of velocity -ve. Similarly, do of the y components of velocity. The coordinate system decides the signs of the displacement, velocity and acceleration.

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A tennis ball hits vertically upwards with a velocity of 20m per sec. square, How do I calculate the maximum height reached?

Startrekker

Community Answer

The maximum height attained by an object is given by: Hmax = ((U sinx)^2) /2g, Hmax = 400/20 = 20m, (since sinx = 1, since angle x=90 degree). Tennis ball reaches a height of 20m.

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