Ball thrown from top of building

In summary, the rock was thrown from the top of a 100-ft building at an unknown angle above the horizontal. It struck the ground 160 ft away from the base of the building after 5.0 seconds. Using the given information, the x component of the initial velocity was calculated to be 32 ft/s. The equation for maximum height does not seem to be helpful in this case. Instead, the vertical displacement at a particular time can be used to determine the initial vertical velocity.
  • #1
jhoffma4
5
0
rock is thrown from the edge of the top of a 100-ft tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160 ft from the base of the building 5.0 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.

I found the x component of the initial velocity to be 32m/s and I think I need to find the y(max). I found t(max) for height to be v(0)/g (for the y component). And y(tmax)=h+1/2(v(o)/g), but this information does not seem to help me any.

Help?
 
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  • #2
Hi jhoffma4,

jhoffma4 said:
rock is thrown from the edge of the top of a 100-ft tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 160 ft from the base of the building 5.0 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.

I found the x component of the initial velocity to be 32m/s

Be careful with your units; this will be 32 ft/s.

and I think I need to find the y(max). I found t(max) for height to be v(0)/g (for the y component). And y(tmax)=h+1/2(v(o)/g)

This equation does not look correct to me.

, but this information does not seem to help me any.

Help?

I would not worry about the maximum height. You know the vertical displacement at a particular time after launch, right? That's all you need to find the initial vertical velocity. What do you get?
 
  • #3


Based on the information provided, we can use the equations of motion to determine the initial velocity of the rock. We know that the horizontal distance traveled by the rock is 160 ft and the time it took to reach the ground is 5.0 seconds. Using the equation d = vt, we can calculate the horizontal component of the initial velocity to be 32 ft/s.

To find the vertical component of the initial velocity, we can use the equation v = u + at, where v is the final velocity (which is 0 ft/s at the highest point), u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time taken to reach the highest point. Since the rock is thrown from the top of a 100-ft tall building, the vertical displacement is 100 ft. Thus, we can write the equation as 0 = u + (-9.8 m/s^2)(t). Solving for u, we get the vertical component of the initial velocity to be 49 m/s.

To find the total initial velocity, we can use the Pythagorean theorem: v = √(vx^2 + vy^2). Plugging in the values we calculated for the horizontal and vertical components, we get the initial velocity of the rock to be approximately 56 m/s.

In conclusion, the rock was thrown with an initial velocity of approximately 56 m/s. It is important to note that this calculation assumes ideal conditions and does not take into account air resistance or other external factors that may have affected the trajectory of the rock.
 

What is the initial velocity of the ball?

The initial velocity of the ball is determined by the speed at which it is thrown from the top of the building. It can also be calculated using the equation v = √(2gh), where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the building.

How long does it take for the ball to reach the ground?

The time it takes for the ball to reach the ground depends on the height of the building and the initial velocity of the ball. It can be calculated using the equation t = √(2h/g), where t is the time, h is the height of the building, and g is the acceleration due to gravity.

What is the maximum height reached by the ball?

The maximum height reached by the ball is equal to the height of the building. This is because the ball will continue to rise until its velocity becomes zero, and then it will start to fall back towards the ground due to the force of gravity.

Does the air resistance affect the motion of the ball?

Yes, the air resistance does affect the motion of the ball. It causes the ball to slow down as it falls towards the ground, reducing its overall velocity. However, for most practical purposes, the effect of air resistance can be ignored unless the ball is thrown from a very high building or is in the air for a long time.

What is the final velocity of the ball when it hits the ground?

The final velocity of the ball when it hits the ground is equal to the initial velocity, but in the opposite direction. This is due to the fact that the ball experiences a constant acceleration of 9.8 m/s² due to gravity, and therefore its velocity will decrease until it becomes zero at the ground.

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