Question

An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the level ground 150m downrange from the launch point throws an apple straight up with the minimum intial speed necessary to meet the path of the arrow.
(a) what is the initial speed of the apple?
(b) at what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

Answer 1

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t  =  5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

                                             = 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s