A ball is thrown upwards with an initial velocity of 70 m/s

1. What is the velocity after 3 seconds?

2. What is the acceleration at the top of its path?

3. How many seconds does it take the ball to go up?

4. How many seconds does it take to hit the ground?

5. How high does the ball go?

1. V = Vo + g*t, g = -9.8 m/s^2.

2. a = g = -9.8 m/s^2.

3. V = Vo + g*Tr, V = 0, Vo = 70 m/s, g = -9.8 m/s^2, Tr = Rise time, Solve for Tr.

4. Tf = Fall time, T = Tr+Tf,
Tf = Tr, T = Tr+Tr = 2Tr = Time
required to reach gnd.

5. V^2 = Vo^2 + 2g*h = 0,
h = -(Vo^2)/2g, g = -9.8 m/s^2,
Vo = 70 m/s,

To answer these questions, we will need to analyze the motion of the ball using the equations of motion. The key equations we will use are:

1. Velocity (v) = Initial velocity (u) + Acceleration (a) * Time (t)
2. Displacement (s) = Initial velocity (u) * Time (t) + 0.5 * Acceleration (a) * Time (t)^2

Now let's solve each question step-by-step:

1. Velocity after 3 seconds:
Using the first equation, plug in the given values:
v = u + a * t
v = 70 m/s (initial velocity) - 9.8 m/s^2 (acceleration due to gravity) * 3 s (time)
v = 70 m/s - 29.4 m/s
v = 40.6 m/s

Therefore, the velocity of the ball after 3 seconds is 40.6 m/s.

2. Acceleration at the top of its path:
At the top of its path, the ball momentarily comes to rest, which means its velocity is 0 m/s. So the acceleration at the top is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.

Therefore, the acceleration at the top of the ball's path is approximately 9.8 m/s^2.

3. Time to go up:
To calculate the time it takes for the ball to go up, we need to find the time it takes for the velocity to reach zero. Since the initial velocity is positive (70 m/s) and the acceleration due to gravity is negative (-9.8 m/s^2), the velocity will become zero once again at the same height it was initially thrown from.

Using the first equation, solve for time (t):
0 = 70 m/s (initial velocity) - 9.8 m/s^2 (acceleration due to gravity) * t
9.8 t = 70
t = 70 / 9.8
t ≈ 7.14 seconds

Therefore, it takes approximately 7.14 seconds for the ball to go up.

4. Time to hit the ground:
To calculate the time it takes for the ball to hit the ground, we need to find when the displacement (height) becomes zero.

Using the second equation, solve for time (t):
0 = 70 m/s (initial velocity) * t - 0.5 * 9.8 m/s^2 (acceleration due to gravity) * t^2
0 = 70t - 4.9t^2

This equation is a quadratic equation in the form of at^2 + bt + c = 0, where:
a = -4.9
b = 70
c = 0

Solve this equation using the quadratic formula, and you will find that the two values for t are t = 0 and t = 14.29 seconds. Since the time cannot be negative, we ignore t = 0.

Therefore, it takes approximately 14.29 seconds for the ball to hit the ground.

5. Height reached by the ball:
To calculate the height reached by the ball, we need to find the maximum height it reaches.

Using the second equation, solve for displacement (s) at the top of its path when the velocity is zero:
0 = 70 m/s * t - 0.5 * 9.8 m/s^2 * t^2
0 = 70t - 4.9t^2

Solve this equation using the quadratic formula, and you will find that the two values for t are t = 0 and t = 14.29 seconds. Since the time cannot be negative, we ignore t = 0.

Now plug in the value of t into the displacement equation:
s = 70 m/s * 7.14 s - 0.5 * 9.8 m/s^2 * (7.14 s)^2
s ≈ 249.66 meters

Therefore, the ball reaches a height of approximately 249.66 meters.