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hellokitty1647
13.12.2019 •
Physics
Aprojectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.
using the formulas of the projectile motion, in how many seconds will the projectile strike the ground? (round your answer to the nearest tenth of a second.)
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Ответ:
Explanation:
Initial velocity = 450 ft/s = 137.16 m/s
Angle, θ = 70°
Consider the vertical motion of projectile,
When the projectile return to the ground we have
Displacement, s = 0 m
Acceleration, a = -9.81 m/s²
Initial velocity, u = 137.16 x sin70 = 128.89 m/s
Substituting in s = ut + 0.5 at²
s = ut + 0.5 at²
0 = 128.89 x t + 0.5 x (-9.81) x t²
t² - 26.28 t = 0
t ( t- 26.28) = 0
t = 0 s or t = 26.28 s
After 26.28 seconds projectile returns 26.28 seconds.
Ответ:
Given the distance r = 2/1000 m, the force between them F = 0.0104 N, the mass of the two object can be calculated using formula:
F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2
And where G = is the gravitational constant (6.67E-11 m3 s-2 kg-1)
The mass of the two objects are 24.96 kg