Respuesta :
Answer: 26.7 m/s
Explanation:
The formulae for the maximum height of a projectile motion is given as
H = u²× (sinx)²/2g
Where x is the angle of inclination = 40° and g is the acceleration due to gravity = 9.8m/s².
u = initial velocity of the projectile motion = 30 m/s
By substituting the parameters, we have that
Hmax = 30² × (sin40°)²/2×9.8
Hmax = 900 × 0.4132/ 19.6
Hmax = 371.858/ 19.6 = 18.9m
But the height (H) of the projectile is 50% of the maximum height.
H = 0.5× Hmax
H = 0.5 × 18.9
H = 9.5m
Recall that a projectile motion is defined in the vertical axis by the equation given below.
v² = u² - 2gH
v² = 30² - 2 ×9.8 × 9.5
v² = 900 - 185.93
v² = 714.07
v = √714.07
v = 26.7 m/s
Answer:
26.74 m/s
Explanation:
The vertical and horizontal of the velocity of the projectile when it first fired from ground level at an angle of 40 degrees
[tex]v_v = vsin\theta = 30sin40^o = 19.3 m/s[/tex]
[tex]v_o = vcos\theta = 30cos40^o = 23 m/s[/tex]
As the ball travels, vertically speaking its kinetic energy is converted into potential energy. When it reaches its maximum height the kinetic energy would be 0. Also let potential energy be 0 at around, we have the following energy conservation equation:
[tex]E_k = E_p[/tex]
[tex]mv_v^2/2 = mgH[/tex]
Let g = 9.81 m/s2. We can divide both sides by m
[tex]v_v^2/2 = gH[/tex]
[tex]H = v_v^2/2g = 19.3^2/(2*9.81) = 18.95 m[/tex]
So half of the maximum height is h = H/2 = 9.48 m
Using the same energy equation at h = 9.48 m then we have
[tex]mv_v^2/2 = mv_h^2/2 + mgh[/tex]
Again, divide both sides by m, and multiply by 2
[tex]v_v^2 = v_h^2 + 2gh[/tex]
[tex]v_h^2 = v_v^2 - 2gh = 19.3^2 - 2*9.81*9.48 = 186[/tex]
[tex]v_h = \sqrt{186} = 13.64 m/s[/tex]
If we ignore air resistance, then nothing could affect the horizontal velocity, it would stay the same at h = 9.48 m, 23m/s
[tex]V = \sqrt{v_h^2 + v_o^2} = \sqrt{13.64^2 + 23^2} = \sqrt{186.0496 + 529} = \sqrt{715.0496} = 26.74m/s[/tex]
