Question

a ball is projected from the ground with a spped of 25m\s two sec later it just clear a wall of 5m high.find the angle of projection,max height. how far beyond the ballagain till the ground?

Answer 1

Answer on Question #45288-Physics-Mechanics-Kinematics-Dynamics

A ball is projected from the ground with a speed of $25\mathrm{m/s}$ two sec later it just clear a wall of $5\mathrm{m}$ high. Find the angle of projection, max height. How far beyond the ball again till the ground?

Solution

Let the angle of projection be $\theta$ . Considering vertical motion at $t = 2$

$$5 = 25 \sin \theta \cdot 2 - \frac {1}{2} \cdot 9.8 \cdot 2 ^ {2} \rightarrow \sin \theta = 0.492.$$

The angle of projection is

$$\theta = \sin^ {- 1} 0.492 = 29.5 {}^ {\circ}.$$

The maximal height of projection is

$$H = \frac {(v \sin \theta) ^ {2}}{2 g} = \frac {(25 \cdot 0.492) ^ {2}}{2 \cdot 9.8} = 7.7 \ m.$$

The range of projection is

$$R = \frac {v ^ {2} \sin 2 \theta}{g} = \frac {25 ^ {2} \sin 59}{9.8} = 54.7 \ m.$$