A ball is thrown directly upward with an initial velocity of 12 m/s. If the ball is released from an initial height of 2.9 m above ground, how long is the ball in the air before landing on the ground? Ignore air drag.
height above ground =
Y = 12 t - 4.9 t^2 + 2.9
SWet that eq
height above ground =
Y = 12 t -4.9 t^2 + 2.9 (meters)
Set Y equal to zero and solve for t.
4.9t^2 -12t -2.9 = 0
There will be two solutions. Take the positive one.
Use thye "quadratic formula":
t = (1/9.8)[12 + sqrt(144 + 56.84)]
= 2.671 s
To find the time it takes for the ball to land on the ground, we can use the equations of motion.
Step 1: Analyze the motion of the ball.
The ball is thrown directly upward, so its initial velocity is positive (+12 m/s). Gravity will cause the ball to decelerate until it reaches its highest point, and then accelerate downward until it lands on the ground.
Step 2: Determine the key values.
The initial velocity (u) of the ball is +12 m/s.
The initial height (h) of the ball above the ground is +2.9 m.
The acceleration due to gravity (g) is approximately -9.8 m/s^2 (negative because it acts in the opposite direction of the initial velocity).
Step 3: Determine the time it takes for the ball to reach its highest point.
At the highest point, the velocity of the ball is zero. Using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can solve for t.
0 = 12 - 9.8t
Simplifying the equation, we get:
9.8t = 12
t = 12 / 9.8
t ≈ 1.22 seconds
Step 4: Determine the total time for the ball to land on the ground.
To find the total time, we multiply the time it takes for the ball to reach its highest point by 2, since the time taken to reach the highest point is the same as the time taken to fall back to the ground.
Total time = 2 * t
Total time = 2 * 1.22
Total time ≈ 2.44 seconds
Therefore, the ball will be in the air for approximately 2.44 seconds before landing on the ground.