Respuesta :
Answer:
62.4m
Explanation:
a) Horizontal distance traveled = x = [tex]v_x[/tex] * t
where,
[tex]v_x[/tex] = horizontal velocity
and t = time in the air)
Time in the air t can be solved using the equation for y:
[tex]i_y=y_o+v_o_yt - 0.5gt^2[/tex]
where, [tex]y_o[/tex] = initial height i.e 0.85 m
[tex]v_o_y[/tex] = initial vertical velocity
and g = 9.8 m/s^2
When y = 0, the dog has hit the water.
So set y = 0 and solve for t.
[tex]v_o_y[/tex] = 8.62 m/s [tex]\times[/tex] (sin 28) = 4.04 m/s
0 = 0.85 m + (4.04 m/s)t - 0.5gt²
0.5t²-4.04t-0.85
Solve this quadratic formula for t: the solutions are t = 8.2 s and t = -0.2. Reject the negative solution, so t = 8.2 s.
How far does the dog travel horizontally in 9.2 s?
x = [tex]v_x[/tex] * t = (8.62 m/s) [tex]\times[/tex] (cos 28) [tex]\times[/tex] 8.2 s = 62.4 m
She travels the horizontal distance before hitting the surface of the water is 62.4m
What is projectile?
When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion.
a) Horizontal distance traveled = x = V(x) * t
where, V(x) is the horizontal velocity and t = time in the air
Time in the air t can be solved using the equation for y:
hy =yo + Voy x t - 1/2gt²
where, yo= initial height i.e 0.85 m and Voy = initial vertical velocity
and g = 9.8 m/s^2
When y = 0, the dock has hit the water.
So set y = 0 and solve for t.
Voy = 8.62 m/s x (sin 28) = 4.04 m/s
Substitute the values, we get time as
0 = 0.85 m + (4.04 m/s)t - 0.5gt²
0.5t²-4.04t-0.85 =0
Solving the quadratic formula for t, we have
time t = 8.2 s and t = -0.2.
As time can't be negative, so, t = 8.2 s.
The horizontal distance is
x = Vo * t = (8.62 m/s) (cos 28) 8.2 s = 62.4 m
Thus, she travels 62.4 m.
Learn more about projectile.
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