Question

While standing on an open bed of a truck moving at 35 m/s, an archer sees a duck flying directly overhead. The archer shoots an arrow straight up at the duck and misses. The arrow leaves the bow with a vertical velocity of 98 m/s.
a.) How long does it remain in the air?
b.) The truck maintains a constant speed of 35 m/s and does not change its direction. What horizontal distance does the arrow travel while it is in the air?
c.) Where does the arrow finally land (ignore air resistance)?

Answer 1

Answer:

Part a)

$t = 20 s$

Part b)

$x = 700 m$

Part c)

Since horizontal speed of truck and the arrow is same so the arrow will strike at the position of the archer as they both moving with same speed

Explanation:

Part a)

As we know that the arrow moves with uniform acceleration in vertical direction so we can use kinematics in Y direction

Here we know that

$\Delta y = v_y t + \frac{1}{2}at^2$

since the arrow lands at the same height so its vertical displacement of whole motion is zero

so we will have

$0 = 98 t - \frac{1}{2}(9.8) t^2$

$0 = 98 t - 4.9 t^2$

$t = 20 s$

Part b)

Horizontal distance moved by the arrow

$x = v_x t$

here horizontal speed of arrow is same as that of speed of truck

so we will have

$x = 35 \times 20$

$x = 700 m$

Part c)

Since horizontal speed of truck and the arrow is same so the arrow will strike at the position of the archer as they both moving with same speed