A ball is thrown upward with an initial velocity of 6.0m/s by someone on top of the building 34m tall who is leaning over the edge so that the ball will not strike the building on the return trip. (a) How far above the ground will the ball be at the end of 1.0s? (b) What is the ball's velocity at the time? (c) When and with what speed will the ball strike the ground?
a. hf=hi+vi*t-4.9t^2 hi=34 vi=6, t=1
b. vf=vi-9.8t
c. hf=hi+ same as a but now hf=0, solve for t.
To find the answers to these questions, we can use the equations of motion for an object in freefall with constant acceleration due to gravity. The key equations we'll use are:
1. Vertical displacement (height): h = ut + (1/2)gt^2
2. Final velocity: v = u + gt
Where:
- h is the vertical displacement (height)
- u is the initial velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time
(a) How far above the ground will the ball be at the end of 1.0s?
To find the height at the end of 1.0s, we need to substitute the values into the equation. Here's how we can do it:
- u = 6.0 m/s (initial velocity)
- t = 1.0 s (time)
Using the formula: h = ut + (1/2)gt^2
h = (6.0 m/s)(1.0 s) + (0.5)(9.8 m/s^2)(1.0 s)^2
h = 6.0 m + 4.9 m
h = 10.9 m
Therefore, at the end of 1.0s, the ball will be approximately 10.9 meters above the ground.
(b) What is the ball's velocity at that time?
To find the ball's velocity at 1.0s, we can use the equation v = u + gt.
- u = 6.0 m/s (initial velocity)
- t = 1.0 s (time)
- g = 9.8 m/s^2 (acceleration due to gravity)
Using the formula: v = u + gt
v = 6.0 m/s + (9.8 m/s^2)(1.0 s)
v = 6.0 m/s + 9.8 m/s
v = 15.8 m/s
Therefore, the ball's velocity at 1.0s is approximately 15.8 m/s.
(c) When and with what speed will the ball strike the ground?
To find when the ball will strike the ground and its speed at that time, we need to consider the vertical displacement (height). The ball will hit the ground when its height (h) is equal to zero.
Using the formula: h = ut + (1/2)gt^2
0 = (6.0 m/s)t + (0.5)(9.8 m/s^2)t^2
This equation is a quadratic equation, we can solve it by factoring, completing the square, or using the quadratic formula. The quadratic equation leads to two values of time, but we only need the positive value since time cannot be negative.
After solving the equation, we find that t ≈ 1.24s or t ≈ 5.00s. However, the time t = 5.00s is for the entire motion (up and down), and we're interested in the time for the ball to hit the ground. Therefore, we consider t = 1.24s as the time for the ball to hit the ground.
To calculate the speed at that time, we can use the formula: v = u + gt
- u = 6.0 m/s (initial velocity)
- t = 1.24 s (time)
- g = 9.8 m/s^2 (acceleration due to gravity)
Using the formula: v = u + gt
v = 6.0 m/s + (9.8 m/s^2)(1.24 s)
v = 6.0 m/s + 12.2 m/s
v ≈ 18.2 m/s
Therefore, the ball will strike the ground at approximately 1.24 seconds after being thrown, with a speed of around 18.2 m/s.