A baseball is hit straight up into the air at an initial velocity of 35 m/s. How long does it take for the
baseball to reach its highest position?
What is the speed of a water bomb after 1.75 s if it is released from rest at a height of 110m above the ground and how high off the ground is the ball?
baseball: 35-9.81t = 0
water bomb: v = -9.81t
h = 110 - 4.9t^2
To find the time it takes for the baseball to reach its highest position, we can use the kinematic equation for vertical motion. The equation is:
v_f = v_i + a*t
Where:
v_f is the final velocity (0 m/s at the highest position)
v_i is the initial velocity (35 m/s)
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time we want to find
Rearranging the equation to solve for time, we have:
t = (v_f - v_i) / a
Plugging in the values, we get:
t = (0 - 35) / -9.8
t = 3.57 seconds (rounded to two decimal places)
So, it takes approximately 3.57 seconds for the baseball to reach its highest position.
To find the speed of a water bomb after 1.75 seconds if it is released from rest at a height of 110m above the ground, we can use another kinematic equation. The equation is:
v_f = v_i + a*t
Where:
v_f is the final velocity (the speed we want to find)
v_i is the initial velocity (0 m/s as it is released from rest)
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time (1.75 seconds)
Plugging in the values, we have:
v_f = 0 + (-9.8) * 1.75
v_f = -17.15 m/s (rounded to two decimal places)
The speed of the water bomb after 1.75 seconds is approximately 17.15 m/s upwards (assuming upwards is positive).
As for the height of the ball above the ground, without knowing the initial velocity at release, it is not possible to determine the height solely based on the given information. More information, such as the overall trajectory or additional data, would be needed to accurately calculate the height off the ground.