Asked by Michael
A shell is fired from the ground with an initial speed of 1.61 multiplied by 103 m/s at an initial angle of 39° to the horizontal.
(a) Neglecting air resistance, find the shell's horizontal range.
(b) Find the amount of time the shell is in motion.
For the first one I tried to find the horizantle component by using 1.6X10^3 Cos 39 and i got some huge number..
(a) Neglecting air resistance, find the shell's horizontal range.
(b) Find the amount of time the shell is in motion.
For the first one I tried to find the horizantle component by using 1.6X10^3 Cos 39 and i got some huge number..
Answers
Answered by
drwls
(a) The horizontal range is
2 (Vo^2) sin 39 cos 39/g
= Vo^2 sin 78/g
= 2.584*10^5 m = 258.4 km
(b) The time in the air is
T = 2 Vo sin 39/g = 207 seconds
2 (Vo^2) sin 39 cos 39/g
= Vo^2 sin 78/g
= 2.584*10^5 m = 258.4 km
(b) The time in the air is
T = 2 Vo sin 39/g = 207 seconds
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