Question

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an initial speed of v0 = 346 m/s. What is the horizontal distance covered by the shell after 5.03 s of flight? Submit Answer Tries 0/12 What is the height of the shell at this moment?

Answer 1

Answer:

1317.4 m

Explanation:

We are given that

Angle=$\alpha=40.8^{\circ}$

Initial speed =$v_0=346m/s$

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=$v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed=$v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance =$Horizontal\;velocity\times time$

Using the  formula

Horizontal distance=$261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m