How long does it take for the ball to hit the ground 7 meters below if it's thrown straight up wit intial velocity of 2 m/s squared?
Solve this equation for time t:
Vo*t -(g/2)*t^2 = 2t - 4.9 t^2 = -7
can you show me a little more work please I am understanding where you are headed?
I wrote down the equation for height (above the release point) vs. time. It ends up 7 meters below the release point.
Solve the quadratic equation that I set up for you. It will lead to the answer.
You will have to a bit of algebra.
got it thanks.
To determine the time it takes for the ball to hit the ground, we can break down the problem into two phases: the ascent and descent phases.
In the ascent phase, the ball is thrown upwards with an initial velocity of 2 m/s. At the highest point, the velocity becomes 0 m/s before it starts descending.
To find the time taken for the ascent phase, we can use the formula:
time = (final velocity - initial velocity) / acceleration
Since the final velocity is 0 (at the highest point), the acceleration is -9.8 m/s² (due to the force of gravity pulling the ball downward).
time = (0 - 2) / -9.8
time = -2 / -9.8
time = 0.204 seconds (rounded to three decimal places)
Next, we can calculate the time it takes for the ball to fall from the highest point to the ground, considering that the ball is initially 7 meters above the ground.
We can use the formula for distance traveled during free fall:
distance = initial velocity * time + (0.5 * acceleration * time^2)
Since the initial velocity when falling is 0 m/s, the distance is 7 meters, and the acceleration remains -9.8 m/s².
7 = 0.5 * -9.8 * time^2
Rearranging the equation:
0.5 * -9.8 * time^2 = 7
Simplifying:
-4.9 * time^2 = 7
Dividing both sides by -4.9:
time^2 = -7 / -4.9
time^2 ≈ 1.43
Taking the square root of both sides:
time ≈ √1.43
time ≈ 1.196 seconds (rounded to three decimal places)
Now, to find the total time taken, we add the time for the ascent and the time for the descent:
total time = time for ascent + time for descent
total time ≈ 0.204 + 1.196
total time ≈ 1.4 seconds (rounded to one decimal place)
Therefore, it takes approximately 1.4 seconds for the ball to hit the ground, 7 meters below, when thrown straight up with an initial velocity of 2 m/s.