Answer:
(a) - 165.032 m/s
(b) 238.37 m/s
Explanation:
initial horizontal velocity, ux = 172 m/s
height, h = 1390 m
g = 9.8 m/s^2
Let it strikes the ground after time t.
Use second equation of motion in vertical direction
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
-1390 = 0 - 0.5 x 9.8 x t^2
t = 16.84 second
(a) Let vy be the vertical component of velocity as it strikes the ground
Use first equation of motion in vertical direction
vy = uy - gt
vy = 0 - 9.8 x 16.84
vy = - 165.032 m/s
Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.
(b)
The horizontal component of velocity remains constant throughout the motion.
vx = 172 m/s
vy = - 165.032 m/s
The resultant velocity is v.
[tex]v=\sqrt{172^{2}+165.032^{2}}[/tex]
v = 238.37 m/s
Thus, teh velocity with which it hits the ground is 238.37 m/s.
