Bob jumps from a 2.0 m high table and wants to jump over a 0.85 m radius ball. How fast must Bob move horizontally to not land on the ball?
If we use kinematics equations for projectile, the answer is 2.66 m/s; however, with this velocity the person will land on the ball. I need to find the velocity at which the person will just miss the ball. And I know this velocity is less than 5 m/s. Thank you for your help.
vertical motion
v = 9.81 t
y = 2 - .5(9.81)t^2
horizontal motion
v = u which is constant
x = u t
parabola
so t = x/u
y = 2 - 4.9 x^2/u^2
now the ball.
You did not tell me where the center of the ball is on the floor. I will assume it is at x = .85 and y = .85
If we just want to clear the top of the ball then the trajectory passes through the point (.85,1.7)
1.7 = 2 - 4.9 (.85)^2/u^2
-.3 = - 3.54/u^2
u = 3.44 m/s
Of course I should check if the Bob hits the ball further down on the far side after clearing the top. You can do that by solving for any other intersection of Bob and ball at x>.85
Hi Damon, thank you for your answer. With u=3.44m/s, Bob will hit the ball on the far side. The instructor showed a graph of the trajectory in class. So we know u is greater than 3.44 m/s but smaller than 5 m/s (he did a trajectory graph for u=5 m/s and Bob passed the ball by a distance). What he wants is an exact u that will help Bob not hit the ball.
To find the velocity at which Bob will just miss the ball, we can use the concept of projectile motion and apply the equations of motion. Let's break down the problem step by step:
1. Firstly, let's find the time it takes for Bob to reach the maximum height by using the equation for vertical motion:
* Initial vertical velocity (u) = 0 (as Bob jumps horizontally)
* Vertical displacement (s) = 2.0 m
* Acceleration due to gravity (g) = 9.8 m/s^2
Using the equation s = ut + (1/2)gt^2, we can rearrange it to t = sqrt(2s/g). Substituting the values, we find t = sqrt(2 * 2.0 / 9.8) = 0.64 s.
2. Now, let's find the horizontal distance Bob travels during this time. Since Bob wants to just miss the ball, the horizontal distance traveled should be equal to or greater than the diameter of the ball, which is twice the radius (2 * 0.85 m = 1.70 m).
* Horizontal distance (d) = ?
* Horizontal velocity (v_x) = ?
Using the equation d = v_x * t, we can rearrange it to v_x = d / t. Substituting the values, we have v_x = 1.70 m / 0.64 s ≈ 2.66 m/s.
3. We know that with this velocity, Bob will land on the ball. So, we need to adjust the velocity to ensure Bob just misses the ball.
To find the maximum acceptable horizontal velocity, we can consider the time it takes for the ball to fall under gravity from the height Bob jumps down and then adjust Bob's velocity accordingly. Let's calculate that:
* Vertical displacement of the ball (s_ball) = 2.85 m (2.0 m + 0.85 m)
Using the same equation as before, t_ball = sqrt(2 * 2.85 / 9.8) ≈ 0.80 s.
4. Next, we find the horizontal distance the ball travels during this time:
* Horizontal distance traveled by the ball (d_ball) = v_ball * t_ball
We already know that v_ball should be less than 5 m/s. Therefore, d_ball < 5 m/s * 0.80 s = 4 m.
To ensure Bob just misses the ball, his horizontal distance traveled (d) should be greater than d_ball + ball radius = 4 m + 0.85 m = 4.85 m.
Therefore, Bob's required horizontal velocity should be greater than d / t, which is greater than 4.85 m / 0.64 s ≈ 7.58 m/s.
5. However, you mentioned that the required velocity is less than 5 m/s. This means it is not possible for Bob to jump over the ball without landing on it, given the initial conditions.
In conclusion, the velocity required for Bob to just miss the ball is greater than 7.58 m/s, which contradicts the provided constraint of the velocity being less than 5 m/s. Therefore, Bob cannot achieve his desired outcome with the given conditions.