Question

A rabbit leaps into the air at an angel of 30 degrees and lands back on the ground 8.00m away in a time of 1.50s. What is the maximum height the rabbit has reached?

Answer 1

Answer on Question #43406, Physics, Mechanics | Kinematics | Dynamics

A rabbit leaps into the air at an angle of 30 degrees and lands back on the ground $8.00\mathrm{m}$ away in a time of 1.50s. What is the maximum height the rabbit has reached?

Solution:

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown near the earth's surface, and it moves along a curved path under the action of gravity only.

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other.

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration due to gravity is constant $(g = 9.81\mathrm{m} / \mathrm{s}^2)$.

Equations related to trajectory motion are given by

Horizontal distance, $x_{max} = v_{0x}t$

Vertical distance, $y = y_{0} + v_{0y}t - \frac{1}{2} gt^{2}$

Horizontal range, $R = x_{max} = \frac{v_0^2\sin 2\theta}{g}$

Maximum height reached, $H = \frac{v_0^2\sin^2\theta}{2g}$

where $v_{0}$ is the initial velocity.

From third equation we have the initial velocity

$$v _ {0} = \sqrt {\frac {R g}{\sin 2 \theta}} = \sqrt {\frac {8 . 0 \cdot 9 . 8 1}{\sin 6 0 {}^ {\circ}}} = 9. 5 2 \mathrm {m} / \mathrm {s}$$

From fourth equation

$$H = \frac {v _ {0} ^ {2} \sin^ {2} \theta}{2 g} = \frac {9 . 5 2 ^ {2} \sin^ {2} 3 0 {}^ {\circ}}{2 \cdot 9 . 8 1} = 1. 5 5 5 \approx 1. 6 \mathrm {m}$$

Answer: $H = 1.6 \, \text{m}$

http://www.AssignmentExpert.com/