Question

Calculate the maximum horizontal distance travelled by a ball thrown with a velocity
Of 40m/s without hitting the ceiling of an auditorium of height 25m.

Answer 1

Question 32221

Maximum height is achieved at the half of time of movement. At this point

$h_{max} = \frac{v_0^2\sin^2\theta}{2g}$ , where $\theta$ is the angle at which object was thrown. If maximum possible height is $25m$, then using last formula it is possible to find the angle under which ball must be thrown not to hit the ceiling (25m).

$h_{max} = \frac{v_0^2\sin^2\theta}{2g} \Rightarrow \sin^2\theta = \frac{2gh_{max}}{v_0^2} = 0.3125$ , which gives the angle

$\sin \theta = 0.56, \theta = 34.05$ degrees .

The horizontal distance is $l = \frac{v_0^2}{g} \sin 2\theta = 148.45 \, m$ .