Question

a ball is thrown horizontally from the top of a tower with a velocity of 40m/s. Take g=10m/s. find the horizontal abd vertical displacement after 1, 2, 3, 4, 5 seconds. plot the path of motion of the ball. ​

Answer 1

Answer:

hope it will help u in solving

Answer 2

Answer:

The values of all horizontal and vertical displacements are as below

$(s _{x}) _{1 } = 40 \times 1 = 40m \\ (s _{x}) _{2 } = 40 \times 2 = 80m \\ (s _{x}) _{3} = 40 \times 3 = 120m \\ (s _{x}) _{4} = 40 \times 4 = 160m \\ (s _{x}) _{5} = 40 \times 5 = 200m$

$(s _{y}) _{1} = 35.09m \\ (s _{y}) _{2} = 60.38m \\ (s _{y}) _{3} = 75.85m \\ (s _{y}) _{4} = 81.52m \\ (s _{y}) _{5} = 77.37m$

Explanation:

Given that,

The ball is thrown horizontal from top of tower with velocity=40m/s

The horizontal displacements can be found using relation

$s _{x} = = ut$

The horizontal displacement after 1 sec is

$(s _{x}) _{1} = 40 \times 1 = 40m$

Similarly

$(s _{x}) _{2 } = 40 \times 2 = 80m \\ (s _{x}) _{3} = 40 \times 3 = 120m \\ (s _{x}) _{4} = 40 \times 4 = 160m \\ (s _{x}) _{5} = 40 \times 5 = 200m$

The vertical displacement is given by relation

$s _{y} = ut - \frac{1}{2} gt {}^{2}$

Therefore,vertical displacement after 1sec is

$(s _{y}) _{1} = 40 \times 1 - \frac{1}{2} \times 9.81 \times 1 {}^{2} = 35.09m$

Similarly,

$(s _{y}) _{2} = 40 \times 2 - \frac{1}{2} \times 9.81 \times 2 {}^{2} = 60.38m \\ (s _{y}) _{3} = 75.85m \\ (s _{y}) _{4} = 81.52m \\ (s _{y}) _{5} = 77.37m$

Hence the values of all required horizontal and vertical displacements are obtained.The plot is a parabola

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