Question

A catapult launches a large stone at a speed of 45.0 m/s
at an angle of 55.0° with the horizontal. What maximum
height does the stone reach?
(a) 45.7 m (b) 32.7 m (C) 69.3 m (d) 83.2 m (e) 102 m

Answer 1

Answer:

The first step to assess the problem is to do the following:

Explanation:

First thing first is to write the givens:

$v_0=45.0 \frac{\text{m}}{\text{s}}\\\theta=55.0^\circ\\g=9.81\frac{\text{m}}{\text{s}^2}$

The gravitational constant is given typically as 9.81 or 10.

The following step would be to pick out which equation which will solve for the maximum height is going to use which is the following:

$h=\frac{v^2\sin^2(\theta)}{2g}$

The rest is plug and chug make sure your calculator is in degrees:

$h=\frac{(45.0)^2(\sin^2(55.0^\circ)}{2*9.81}\\h=69.255\approx69.3 \ \text{meters}$

Answer 2

The maximum height reached by the stone is 69.3 m. Hence, option (C) is correct.

Given data:

The initial launching speed is, u = 45.0 m/s.

The angle of inclination with respect to horizontal is, α = 55.0°.

The given problem is based on the Projectile motion. The Maximum vertical distance covered by an object during the projectile motion is known as Maximum height. The mathematical expression for the maximum height achieved during the projectile is given as,

$H = \dfrac{u^{2}sin^{2} \alpha}{2g}$

here, g is the gravitational acceleration.

Solving as,

$H = \dfrac{45.0^{2} \times sin^{2} 55.0}{2 \times 9.8}\\\\H \approx 69.3 \;\rm m$

Thus, we can conclude that the maximum height reached by the stone is 69.3 m.

Learn more about the projectile motion here:

https://brainly.com/question/11049671