If the time (up and down) the ball remains in the air is 2.14 s, find its speed when it caught. The acceleration of gravity is 9.8 m/s^2
. Neglect air resistance.
Answer in units of m/s
If the time the ball remains in the air is 2.27 s, find the maximum height hmax the ball attained while in the air.
Answer in units of m.
I honesty don't even know where to begin or what formula to use. (Can kinematic equations even be used?)
To solve these problems, you can use the formulas of motion under constant acceleration. These formulas, known as the kinematic equations, can be used to calculate various quantities such as velocity, displacement, and time.
For the first question regarding the speed of the ball when it is caught, we know the time the ball remains in the air (2.14 s) and the acceleration due to gravity (9.8 m/s^2). We need to find the speed at that moment.
The formula you can use to solve this problem is:
v = u + at
where:
v = final velocity (speed when the ball is caught)
u = initial velocity (speed when the ball was thrown, assumed to be zero when it is thrown upwards)
a = acceleration due to gravity (negative value as it acts in the opposite direction of motion)
t = time the ball remains in the air
Since the initial velocity is zero, the term 'u' becomes zero. We can substitute the known values into the formula:
v = 0 + (-9.8 m/s^2) * 2.14 s
Simplifying this equation gives us:
v = -9.8 m/s^2 * 2.14 s = -20.972 m/s
The negative sign indicates that the velocity is in the opposite direction of the initial throw, which is expected as the ball is caught. However, since velocity is a vector quantity, we typically ignore the negative sign when considering only the magnitude. Therefore, the speed of the ball when it is caught is approximately 20.972 m/s.
Moving on to the second question regarding the maximum height hmax attained by the ball, we are given the time the ball remains in the air (2.27 s) and the acceleration due to gravity (9.8 m/s^2).
The formula you can use to determine the maximum height is:
hmax = ut + (1/2)at^2
where:
hmax = maximum height
u = initial velocity (the speed when the ball was thrown, assumed to be zero when it is thrown upwards)
a = acceleration due to gravity (negative value as it acts in the opposite direction of motion)
t = time the ball remains in the air
Again, since the initial velocity is zero, the 'u' term becomes zero:
hmax = 0 * 2.27 s + (1/2)(-9.8 m/s^2)(2.27 s)^2
Simplifying this equation gives us:
hmax = (1/2)(-9.8 m/s^2)(2.27 s)^2 = -24.9083 m^2/s^2
Since the height cannot be negative, we ignore the negative value and take the magnitude:
hmax ≈ 24.9083 m
Therefore, the maximum height attained by the ball is approximately 24.9083 meters.