A ball is thrown straight up with an initial speed of 60 m/s. How high does it go? Assume the accelera- tion of gravity is 10 m/s22
H=u^2/2g H=60^2/20=3600/20=180m
To calculate the maximum height reached by the ball, we can use the kinematic equations of motion. The given information includes the initial speed of the ball (60 m/s) and the acceleration due to gravity (10 m/s^2).
The first step is to determine the time it takes for the ball to reach its highest point. We can use the equation:
v_f = v_i + at
where v_f is the final velocity (which is 0 m/s at the highest point), v_i is the initial velocity (60 m/s), a is the acceleration (-10 m/s^2 due to gravity), and t is the time.
Rearranging the equation, we have:
t = (v_f - v_i) / a
Plugging in the values:
t = (0 - 60) / -10 = 6 seconds
So, it takes 6 seconds for the ball to reach its highest point.
Next, we can calculate the maximum height reached using the equation:
h = v_i * t + (1/2) * a * t^2
where h is the maximum height, v_i is the initial velocity (60 m/s), a is the acceleration (-10 m/s^2), and t is the time (6 seconds).
Plugging in the values:
h = 60 * 6 + (1/2) * (-10) * (6)^2 = 180 meters
Therefore, the ball reaches a maximum height of 180 meters.
v = Vi + a t
0 = 60 -10 t because v = 0 at top
t = 6 seconds at top
then
h = 0 + Vi t +(1/2)a t^2
h = 60 (6) - 5 (36)
h = 360 - 180 = 180 meters
or simply average speed up = 60/2 = 30 m/s
30 m/s * 6 seconds = 180 meters