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Anika [276]
3 years ago
9

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

Physics
1 answer:
pav-90 [236]3 years ago
5 0
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

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.

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With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
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The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

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Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

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\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

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\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

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Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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