What Is the Final Speed of a Steel Ball Launched from a Platform?

[rqu1ntana]

A steel ball is launched at 25 degrees above horizontal from the top of a platform that is 15 meters above ground. If the initial speed is 10m/sec, what is the final speed when it strikes the ground?

 

[tiny-tim]

hi rqu1ntana! welcome to pf!

show us what you've tried and where you're stuck, and then we'll know how to help!

(use the standard constant acceleration equations, in the x and y directions separately)

 

[rqu1ntana]

The problem is that its for a test and i just want to be sure what i did was right. This is what i did,

I used the formula V_f²=V_i²-2g(y-y_i), where V_i² is the initial velocity, g is gravity=9.8, y is 0, and y_i is the height.

from all the data given this is what it looks like, V_f=(10²-(2*9.8*(-15)))^1/2

The result i get from this is 13.56m/sec, i this right?

 

[tiny-tim]

hi rqu1ntana!

from all the data given this is what it looks like, V_f=(10²-(2*9.8*(-15)))^1/2

The result i get from this is 13.56m/sec, i this right?

yes that looks fine … you've used conservation of energy (so the 25° doesn't matter)

… except that you keyed in -110 instead of 100 !

 

[Nickie]

To find the final speed of the steel ball, we can use the equation for projectile motion: vf = vi + at, where vf is the final speed, vi is the initial speed, a is the acceleration due to gravity, and t is the time the ball takes to reach the ground.

First, we need to find the time the ball takes to reach the ground. This can be calculated using the equation y = viy*t + (1/2)*a*t^2, where y is the vertical displacement, viy is the initial vertical velocity, and a is the acceleration due to gravity. In this case, y = -15m (since the ball starts at a height of 15m and ends at ground level), viy = 10m/s (since the ball is launched at an angle of 25 degrees above horizontal), and a = -9.8m/s^2 (the acceleration due to gravity). Solving for t, we get t = 1.42 seconds.

Now, we can plug this value for t into the first equation to find the final speed: vf = 10m/s + (-9.8m/s^2)(1.42s) = 10m/s - 13.96m/s = -3.96m/s.

Therefore, the final speed of the steel ball when it strikes the ground is approximately 3.96m/s. It is important to note that the negative sign indicates that the ball is moving downwards, as expected.