Question

a batter hits a ball so that it lifts the bat at speed of 30 m\s at angle 53.1 degree.find the position and velocity of ball at T=2 sec

Answer 1

Answer on Question #45287 - Physics - Mechanics | Kinematics | Dynamics

a batter hits a ball so that it lifts the bat at speed of $30\mathrm{m}\backslash \mathrm{s}$ at angle 53.1 degree.find the position and velocity of ball at $T = 2$ sec

Solution:

$\mathrm{V}_0 = 30\frac{\mathrm{m}}{\mathrm{s}} -$ initial velocity of the ball;

$\alpha = 53.1{}^{\circ} - \text{angle of projection};$

$\mathrm{T} = 2\mathrm{s}$ - travelling time;

$\mathrm{V}_{1}$ - final velocity of the ball;

$\mathrm{x}_{1}$ - final horizontal position of the ball;

$\mathrm{y}_{1}$ - final vertical position of the ball;

From the right triangle ABC:

$\mathrm{V_{x0} = V_0\cdot cos\alpha ;V_{y0} = V_0\cdot sin\alpha}$

Equation of motion of the ball along the X-axis:

$$x _ {1} = V _ {x 0} T \cos \alpha = 3 0 \frac {m}{s} \cdot 2 s \cdot \cos 5 3. 1 {}^ {\circ} = 3 6 m$$

Equations of motion along the Y-axis:

$$y _ {1} = V _ {y 0} T - \frac {g T ^ {2}}{2} = 3 0 \frac {m}{s} \cdot 2 s \cdot \sin 5 3. 1 {}^ {\circ} - 9. 8 \frac {m}{s ^ {2}} \cdot \frac {1}{2} \cdot (2 s) ^ {2} = 2 8. 4 m$$

Rate equation for the horizontal component of the final velocity:

$$V _ {1 y} = V _ {0 y} - g T = V _ {0} \cdot \sin \alpha - g T = 3 0 \frac {m}{s} \cdot \sin 5 3. 1 {}^ {\circ} - 9. 8 \frac {m}{s ^ {2}} \cdot 2 s = 4. 4 \frac {m}{s}$$

$V_{1x} = V_{0x}$, because horizontal acceleration is zero.

The final velocity of the ball by the Pythagorean theorem:

$$V_1 = \sqrt{V_{1x}^2 + V_{1y}^2} = \sqrt{\left(4.4 \frac{\mathrm{m}}{\mathrm{s}}\right)^2 + \left(30 \frac{\mathrm{m}}{\mathrm{s}} \cdot \cos 53.1{}^\circ\right)^2} = 18.5 \frac{\mathrm{m}}{\mathrm{s}}$$

Answer: position of the ball: horizontal: $36\,\mathrm{m}$, vertical: $28.4\,\mathrm{m}$; final velocity: $18.5\,\frac{\mathrm{m}}{\mathrm{s}}$.

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