Question

a particle is thrown at some angle with x-axis with speed "u". After 1 second its velocity makes 45 degree with X-axis. After 2 second particle is at its maximum height. Find angle of projection?

Answer 1

a particle is thrown at some angle with x-axis with speed "u". After 1 second its velocity makes 45 degree with X-axis. After 2 second particle is at its maximum height. Find angle of projection?

Solution:

$$t _ {1} = 1 \sec , t _ {2} = 2 \sec$$

X-component of the velocity is constant and equals to $U \cos \alpha$ because the acceleration $\vec{g}$ acts only along the Y-axis.

After 1 second of the flight:

$$\mathrm {X} - \text {a x i s :} U _ {1 x} = U \cos \alpha$$

Rate equation after 1 second of the flight:

$$Y - \text {a x i s :} U _ {1 y} = U \sin \alpha - g t _ {1}$$ $$\tan 4 5 {}^ {\circ} = \frac {U _ {1 y}}{U _ {1 x}} = 1$$ $$(1) \operatorname {a n d} (2) \operatorname {i n} (3): \frac {U \sin \alpha - g t _ {1}}{U \cos \alpha} = 1$$ $$U \sin \alpha - g t _ {1} = U \cos \alpha$$ $$U (\sin \alpha - \cos \alpha) = g t _ {1}$$

After 2 second of the flight Y-component of the velocity is zero. Rate equation after 2 second of the flight:

$$Y - \text{axis: } U \sin \alpha = g t_2$$ $$U = \frac{g t_2}{\sin \alpha}$$

(5)in(4):

$$\frac{g t_2}{\sin \alpha} (\sin \alpha - \cos \alpha) = g t_1$$ $$t_2 \sin \alpha - t_2 \cos \alpha = t_1 \sin \alpha$$

(6) $\div$ $\cos \alpha$

$$t_2 \tan \alpha - t_2 = t_1 \tan \alpha$$ $$\tan \alpha = \frac{t_2}{t_2 - t_1}$$ $$\alpha = \arctan \frac{t_2}{t_2 - t_1} = \arctan \frac{2s}{2s - 1s} = \arctan 2 = 63.43{}^\circ(1.107 \text{ rad})$$

Answer: angle of projection is $\alpha = 63.43{}^\circ(1.107 \text{ rad})$.