Respuesta :
Answer:
a) 3.05 seconds
b) 45.87 m
c) 0.995 seconds
Explanation:
t = Time taken
u = Initial velocity = 30 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = 9.81 m/s² (positive downward, negative upward)
a) Time taken by the ball to reach maximum height
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-30}{-9.81}\\\Rightarrow t=3.05\ s[/tex]
Time taken by the ball to reach maximum height is 3.05 seconds
b) Maximum height
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-30^2}{2\times -9.81}\\\Rightarrow s=45.87\ m[/tex]
Maximum height reached by the ball is 45.87 m
c) If s = 25 m
[tex]v^2-u^2=2as\\\Rightarrow v^2=2as+u^2\\\Rightarrow v^2=2\times -9.81\times 25+30^2\\\Rightarrow v=\sqrt{2\times -9.81\times 25+30^2}\\\Rightarrow v=20.24\ m/s[/tex]
Velocity of the ball at 25 m height is 20.24 m/s
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{20.24-30}{-9.81}\\\Rightarrow t=0.995\ s[/tex]
Time taken by the ball to reach a point 25.0 m above the point of contact between the bat and ball is 0.995 seconds
Answer:
(c) Remember that when you square root, you get two answers. The ball will be at 25 meters two times. Think of a parabola.
Explanation:
if you use yf = yi +vi(t)+ 1/2(a)(t^2),
you'll get -4.9t^2 +30t - 25 = 0, using the quadratic formula you will end up with -30 +/- [tex]\sqrt{410[/tex] all divided by -9.8
t= 1 and 5.1
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