A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 11.0 m/s when the hand is 1.50 m above the ground.
How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
v(final)=v(initial)+at
initial velocity is 15 and final velocity is zero. acceleration is 9.8 right.
how does height fit into this equation. isn't that necessary too.
can u please help me. i keep getting .738s as my answer
height (y) vs time is given by
y = 1.50 + 11 t - 4.9 t^2.
Solve for the time (t) when y = 0. You will have to solve a quadratic equation. Take the positive one of the two answers.
You don't use the velocity vs. time equation in this problem.
To solve this problem, you need to consider the vertical motion of the ball. You can use the kinematic equation for vertical motion:
v(final) = v(initial) + at
Where:
v(final) = final velocity (which is zero in this case, as the ball hits the ground)
v(initial) = initial velocity (11.0 m/s)
a = acceleration (in this case, the acceleration due to gravity, which is -9.8 m/s^2 because it acts downward)
Rearranging the equation, we get:
t = (v(final) - v(initial)) / a
Substituting the given values, we have:
t = (0 - 11.0) / -9.8
t = -11.0 / -9.8
t ≈ 1.12 seconds
The negative signs cancel out, as they only denote the direction. Therefore, the ball is in the air for approximately 1.12 seconds before it hits the ground.
To determine how long the ball is in the air before it hits the ground, we can use the equation for displacement:
s = ut + (1/2)at^2,
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial velocity is 11.0 m/s (upwards), the acceleration is -9.8 m/s^2 (due to gravity), and the displacement is -1.50 m (since the ball is thrown upwards, the displacement is negative). We need to find the time, t.
Rearranging the equation, we have:
s = ut + (1/2)at^2,
-1.50 = (11.0)t + (1/2)(-9.8)t^2.
This equation is a quadratic equation in terms of t. We can solve it by rearranging and using the quadratic formula:
(1/2)(-9.8)t^2 + (11.0)t - 1.50 = 0.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = (1/2)(-9.8), b = 11.0, and c = -1.50.
Plugging in these values, we get:
t = (-11.0 ± √(11.0^2 - 4 * (1/2)(-9.8)(-1.50))) / (2 * (1/2)(-9.8)).
Simplifying this expression, we get:
t = (-11.0 ± √(121.0 + 29.4)) / -9.8.
t = (-11.0 ± √150.4) / -9.8.
t = (-11.0 ± 12.27) / -9.8.
t ≈ 0.238 s, or t ≈ 2.476 s.
Therefore, there are two possible times: approximately 0.238 seconds and approximately 2.476 seconds.
However, since the ball was thrown upwards and the question asks for the time it takes for the ball to hit the ground, we are interested in the positive solution, which is approximately 2.476 seconds.
Therefore, the ball is in the air for approximately 2.476 seconds before it hits the ground.