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Bumek [7]
3 years ago
6

a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity

Physics
1 answer:
stich3 [128]3 years ago
4 0

consider the motion along the X-direction

X = horizontal displacement = 80 m

V_{ox} = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = V_{ox}   t

80 = (v Cos60) (t)

t = 160/v                                         eq-1


consider the motion in vertical direction :

Y = vertical displacement = 20 m

V_{oy}  = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = V_{oy}  t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

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1. 0.42 Hz

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The period of the harmonic motion is equal to the reciprocal of the frequency:

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3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

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- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

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We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

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