Question

A ball is hit at an initial speed of 80 m/s. At what angle should it leave the bat if it has to travel 300 m horizontaly?

Answer 1

Answer on Question #44018-Physics-Mechanics-Kinematics-Dynamics

A ball is hit at an initial speed of $80\mathrm{m / s}$ . At what angle should it leave the bat if it has to travel $300\mathrm{m}$ horizontally?

Solution

The horizontal distance of projectile is

$$d = \frac {v ^ {2}}{g} \sin 2 \theta ,$$

where $v$ is an initial speed of a ball, $g$ is the acceleration of the gravity, $\theta$ is an angle.

So,

$$\sin 2 \theta = \frac {g d}{v ^ {2}} = \frac {9 . 8 \frac {m}{s ^ {2}} \cdot 3 0 0 m}{\left(8 0 \frac {m}{s}\right) ^ {2}} = 0. 4 5 9 3 7 5 \rightarrow \theta = \frac {\sin^ {- 1} 0 . 4 5 9 3 7 5}{2} = 1 3. 7 {}^ {\circ}.$$

Answer: $13.7{}^{\circ}$ .

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