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FrozenT [24]
3 years ago
5

Help please suffering

Physics
1 answer:
inessss [21]3 years ago
3 0

Assuming Lin jumps from and returns to the same level ground, the fact that he stays in the air for 0.8 seconds means it takes him 0.4 seconds to reach his maximum height, at which point his vertical velocity will be 0.

Acceleration is a constant g in the downward direction, and when it's constant it's the same as the average acceleration over any given time interval. Over the 0.4 second interval during which he reaches his maximum height, we have

a_{\rm avg}=-g=\dfrac{0-v_0}{0.4\,\rm s}\implies v_0=3.9\,\dfrac{\rm m}{\rm s}

which is also the same speed he will have when he returns to the ground.

To find the maximum height, we have

0^2-{v_0}^2=2(-g)y_{\rm max}\implies y_{\rm max}=0.78\,\rm m

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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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