Question

an archer fires an arrow with an initial velocity of 4m/sec
atan elevation of 45 degrees.
calculate:
a) the max height attained
b) the time of flight
c) the max range of the arrow

Answer 1

An archer fires an arrow with an initial velocity of $4\mathrm{m / sec}$ at an elevation of 45 degrees. Calculate:

a) the max height attained;

b) the time of flight;

c) the max range of the arrow.

Solution.

$v_{x} = v_{0}\cos \alpha ;$

$v_{y} = v_{0}\sin \alpha .$

a) the max height attained.

$h_{max} = \frac{v_y^2 - v_{0y}^2}{-2g}.$

At the max height $v_{y} = 0$ :

$h_{max} = \frac{-v_{0y}^2}{-2g};$

$h_{max} = \frac{v_0^2}{2g};$

$h_{max} = \frac{v_0^2\sin^2\alpha}{2g}.$

$$h_{max} = \frac{4^2 \cdot \sin^2 45{}^\circ}{2 \cdot 9.8} = 0.4(m).$$

b) the time of flight.

$$h = v_{0y} t - \frac{g t^2}{2};$$ $$h = v_0 \sin \alpha t - \frac{g t^2}{2}.$$

At the end of the flight $h = 0$:

$$0 = v_0 \sin \alpha t - \frac{g t^2}{2};$$ $$v_0 \sin \alpha t = \frac{g t^2}{2};$$ $$v_0 \sin \alpha = \frac{g t}{2};$$ $$t = \frac{2 v_0 \sin \alpha}{g}.$$ $$t = \frac{2 \cdot 4 \cdot \sin 45{}^\circ}{9.8} = 0.58(s);$$

c) the max range of the arrow.

$$l_{max} = v_{0x} t;$$ $$l_{max} = v_0 \cos \alpha t;$$ $$l_{max} = \frac{2 v_0^2 \sin \alpha \cos \alpha}{g}.$$ $$l_{max} = \frac{2 \cdot 4^2 \cdot \sin 45{}^\circ \cdot \cos 45{}^\circ}{9.8} = 1.63(m).$$

Answer:

a) the max height attained: $h_{max} = 0.4m$.

b) the time of flight: $t = 0.58s$.

c) the max range of the arrow: $l_{max} = 1.63m$.