A shell is fired from the ground with an initial speed of 1.51 ✕ 10^3 m/s at an initial angle of 32° to the horizontal. (a) Negl ecting air resistance, find the shell's horizontal range. m (b) Find the amount of time the shell is in motion.
2 answers:
Explanation:
Given that,
Initial speed of the shell,
Angle of projection,
(a) The range of a projectile is given by :
R = 209116.35 meters
(b) Let t is the amount of time the shell is in motion. It can be calculated as :
Here, d = R
t = 138.48 seconds
Hence, this is the required solution.
Answer:
(a) 209.1km
(b) 163.3sec
Explanation:
(a) The range assuming leaving the ground is given by.
d = v2/g x sin 2 angle
d = (1.51*10^3)2/9.8 * sin 2*32..
d = 0.2091*10^6m
d = 209.1km.
(b) The horizontal component of velocity is constant and is equal to
.t = s/v
t = 0.2091*10^6/1.51*10^3*cos 32
t = 163.3sec
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