Hard Kinematics problem: Ball thrown into the air

[Christian Despo]

Homework Statement

A person throws a ball up from a height of 1.5 meters, 3.23 seconds later, the ball hits the ground next to you. Find the maximum height the ball reaches.

Homework Equations

V=V(initial)+at

The Attempt at a Solution

This problem almost seems impossible without knowing what the initial velocity is. And since the can't simply divide the time by 2, I can't find the time it takes to reach the peak.

 

[haruspex]

This problem almost seems impossible without knowing what the initial velocity is.

Consider first the whole trajectory. You know the time, the acceleration and the displacement.
What kinetic equation do you know with those three variables?

Kinematics is something else.

 

[Christian Despo]

Consider first the whole trajectory. You know the time, the acceleration and the displacement.
What kinetic equation do you know with those three variables?

Kinematics is something else.

y=y₀+v₀t+1/2at² and v²=v₀²+2a(x-x₀)

but I am still confused with the steps to solve since I would use the quadratic formula if I were originally given the initial velocity.
Thank you for the reply! I've been trying to solve this for hours.

 

[haruspex]

y=y0+v0t+1/2at2 and v2=v02+2a(x-x0)

The first of those has (implicitly) the three variables I mentioned, but the second does not - it omits time.
In terms of the variables in the first one, what represents displacement?

 

[Christian Despo]

The first of those has (implicitly) the three variables I mentioned, but the second does not - it omits time.
In terms of the variables in the first one, what represents displacement?

Y₀ = 1.5 and y=0 right?

 

[haruspex]

Y₀ = 1.5 and y=0 right?

Yes, so solve the equation to find the launch speed.

 

[PeroK]

Homework Statement

A person throws a ball up from a height of 1.5 meters, 3.23 seconds later, the ball hits the ground next to you. Find the maximum height the ball reaches.

Homework Equations

V=V(initial)+at

The Attempt at a Solution

This problem almost seems impossible without knowing what the initial velocity is. And since the can't simply divide the time by 2, I can't find the time it takes to reach the peak.

To see the problem must have a unique solution you can reason as follows:

If the ball is dropped, it lands on the ground in a short time. Certainly less than 3s. If it is thrown upwards at a low speed, it will take a little longer to land. As you increase the upward speed the ball takes longer to land on ground. At some speed it takes 3.23s. And, if the speed is increased further it takes longer again.

Now, you could actually solve this numerically using a calculator or computer. Keep increasing the initial speed and calculate the time it takes. Eventually you can home in on the initial speed that gives approx 3.23s.

Note that knowing the total time is effectively equivalent to knowing the initial velocity. In either case, the above reasoning shows that the problem must have a unique solution.

 

[Christian Despo]

To see the problem must have a unique solution you can reason as follows:

If the ball is dropped, it lands on the ground in a short time. Certainly less than 3s. If it is thrown upwards at a low speed, it will take a little longer to land. As you increase the upward speed the ball takes longer to land on ground. At some speed it takes 3.23s. And, if the speed is increased further it takes longer again.

Now, you could actually solve this numerically using a calculator or computer. Keep increasing the initial speed and calculate the time it takes. Eventually you can home in on the initial speed that gives approx 3.23s.

Note that knowing the total time is effectively equivalent to knowing the initial velocity. In either case, the above reasoning shows that the problem must have a unique solution.

This helped a lot thank you! I was able to get it!

 

[Christian Despo]

Yes, so solve the equation to find the launch speed.

Thank you for the help I was able to solve it!

 

[neilparker62]

Alternative approach using average velocity (just for interest!)

$$ v_{av}=\frac{Δx}{Δt}=\frac{-1.5}{3.23} $$ Since velocity vs time is linear under constant acceleration: $$v_i+g\frac{Δt}{2}=v_{av}$$ Hence $$v_i=v_{av}-g\frac{Δt}{2}=\frac{Δx}{Δt}-g\frac{Δt}{2} $$ Our point of reference is from where the ball was thrown with upwards positive so g=-9.8ms^-2. If we were also asked for the final velocity of the ball (when it hits the ground): $$v_f=v_{av}+g\frac{Δt}{2}=\frac{Δx}{Δt}+g\frac{Δt}{2} $$