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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Moons gravity is about 1/6 as powerful as it is on Earth, so about 20 pounds.
Explanation:
momentum = mass x velocity
initial momentum = 100 x 15 = 1500kgm/s
after momentum = 100 x 20 = 2000kgm/s
a =(v-u)/t
a = (20-15)/10
a = 5/10
a = 0.5m/s²
f = ma
f = 100 x 0.5
f = 50N