An artillery shell is fired at an angle of 43.7 ◦ above the horizontal ground with an initial speed of 1740 m/s?
An artillery shell is fired at an angle of 43.7
◦
above the horizontal ground with an initial
speed of 1740 m/s.
The acceleration of gravity is 9.8 m/s
2
.
Find the total time of flight of the shell,
neglecting air resistance.
Answer in units of min
Find its horizontal range, neglecting air resistance.
Answer in units of km
More Questions From This User See All
Answers & Comments
Verified answer
Range x = 1740 cos 43.7 x T gives distance in meters. Divide by 1000 for km.
T=2 (1740m/s) Sin 43.7/9.8m/s gives total time in seconds. Divide by 60 for minutes.
The vertical factor of velocity is sine (seventy 3.8°). The horizontal factor of velocity is cosine (seventy 3.8°) Ke = Pe, mgh = ½mv² , gh = ½v², h = v²/2g the optimal altitude is = [[sine (seventy 3.8°)×a million,570m/s]² ÷ 2 × 9.8m/s² s = ½at² , t = ?(2s/a) using the optimal altitude, the time of flight is = ?[ 2 × (max top)/9.8m/s²] × 2 Distance = velocity × time the area traveled horizontally is t (from above) × cosine (seventy 3.8°) × a million,570 m/s
t = 2*V*sinΘ/g sec
R = V²*sin(2Θ)/g meters