Question

A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range R. (a) At what angle θ is the rock thrown? (b) In terms of its original range R, what is the range Rmax the rock can attain if it is launched at the same speed but at the optimal angle for maximum range?(c) Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet?

Answer 1

The rock is thrown at an angle of 45 degrees to achieve a maximum height equal to its horizontal range. The same range is achieved if the rock is launched at the same speed and conditions. However, the optimal launch angle might differ on a planet with a different gravity from Earth's.

Step-by-step explanation:

The question involves the concept of projectile motion and its optimization in Physics. To address the question: (a) A rock is thrown upward from level ground such that its maximum height is equal to its horizontal range R. The angle θ at which the rock is thrown is 45° because, for a given initial velocity and neglecting air resistance, the range is maximal at 45° when the maximum height equals the horizontal range. (b) The maximum range Rmax that the rock can attain if launched at the optimal angle for maximum range, which is again 45°, would be the same as R in this case because the speed and other conditions are unchanged. (c) On a different planet, the value of gravity would affect the projectile's motion; hence the angle for the maximum range might change if gravity is different from Earth's gravity.

Answer 2

a) 76º b) Rmax = 2.1R

Step-by-step explanation:

When reading the exercise you can see that it is a projectile launching exercise, let's write the equations for reach and height.

R = Vo2 sin 2T / g

VY² = Voy² -2gY

The condition they give us

R = Ym

At the point of maximum height the vertical speed is zero (Vy = 0)

0 = Voy2 -2 gY

Y = I'm going 2 / 2g

We match the equations

Voy² /2g = Vo² sin 2θ / g

Vo = Vo sin T

(Vo² sin²θ)/2 = vo² sin 2θ

sin² θ = 2 sin 2θ

Use the relationship. sin 2θ = 2 sin θ cos θ

sin² θ = 2 (2 sin θ cos θ)

sin θ = 4 cos θ

Tan θ = 4

θ = tan⁻¹ 4

θ = 76º

b) to find the maximum range we analyze the expression of the range and see that it will be maximum when the sin 2θ is 1 which implies an angle of 45º, in this case the equation is

Rmax = Vo² /g maximum range

R = Vo²/g sin 2θ

To find the relationship

R / Rmax = sin 2T

Rmax = R / sin (2 76)

Rmax = 2.1 R

c) When reviewing expressions we can see that when equalizing the scope and maximum height the value of the acceleration of gravity is canceled, because the answer does not depend on the severity, consequently, if the experiment is performed in another plan with different acceleration of severity the result is not altered.