Question

A body of mass 5 kg, projected at an angle of 60° from the ground with an initial velocity of 25 m/s, acceleration due to gravity is g = 10 m/s2 , what is the maximum horizontal range covered?​

Answer 1

Answer:

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Answer 2

Answer:

The maximum horizontal range, in this case, is 54.125m.

Explanation:

We know that the formula that we have for calculating the horizontal range is given as below:

R = $v^{2}$sin2θ/g

where we have v = velocity of the projected body

θ is the termed as the angle of projection, and g is the acceleration due to gravity.

From the above-given question, we have the following things given,

m = 5 kg,  θ = 60° , g = 10m/$s^{2}$, v = 25m/s

Therefore putting the given in the equation we have,

$R = 25^{2}$sin120°/10

sin120° = $\frac{\sqrt{3} }{2}$

∴R = 54.125 m

Therefore the maximum horizontal range is found out to be 54.125m.

Hence the answer is 54.125m.