A 5-kg ball is thrown straight up in the air with an initial speed of 24m/s. After its release and before it hits the ground when is the ball in free fall?
My answer is at all times. Is this correct? Thank you.
6 answers
Correct!
1) How long does it take the ball to reach its highest point?
I am not sure what I would divide 24 by to find this.
2) How high above the starting point is the highest point?
I am not sure how to find this.
I am not sure what I would divide 24 by to find this.
2) How high above the starting point is the highest point?
I am not sure how to find this.
Write down the equation for the velocity as a function of time:
V(t) = V(0) - g t
where 9 = 9.81 m/s^2
Here we take the velocity positive if it is in the upward direction. When the ball is at the highest point, the velocity is zero. Clearly if the velocity is not zero, it is either going to move further upward, or if it is negative, it was at a higher point earlier.
V(t) = V(0) - g t
where 9 = 9.81 m/s^2
Here we take the velocity positive if it is in the upward direction. When the ball is at the highest point, the velocity is zero. Clearly if the velocity is not zero, it is either going to move further upward, or if it is negative, it was at a higher point earlier.
Ok but where did the 9 come from?
I know the highest point would be 29.4 but I do not know how to get this.
There are two ways of solution
1. using kinematics.
v =vₒ - g•t,
Velocity at the highest point is zero, =>
0 = vₒ - g•t,
t =vₒ/g.
h = vₒ•t - g•t²/2 = vₒ²/2•g
2. using the law of conservation of energy
KE =PE
m•vₒ²/2 = m•g•h,
h = vₒ²/2•g.
h = (24)²/2•9.8 = 29.4 m
1. using kinematics.
v =vₒ - g•t,
Velocity at the highest point is zero, =>
0 = vₒ - g•t,
t =vₒ/g.
h = vₒ•t - g•t²/2 = vₒ²/2•g
2. using the law of conservation of energy
KE =PE
m•vₒ²/2 = m•g•h,
h = vₒ²/2•g.
h = (24)²/2•9.8 = 29.4 m
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