Will tosses a ball straight upward. Show that if the ball is tossed upward at 18m/s it will be in the air for less than 4s.
Use the height formula.
hf=hi+vo*t-1/2 g t^2 solve for t, you knw hf=hi=0
is there another formula that can be used? I haven't learned that one. Can it be the initial velocity/2
To determine whether the ball will be in the air for less than 4 seconds when tossed upwards at 18 m/s, we can use the equations of motion.
The formula for the time it takes for an object to reach maximum height when thrown vertically upward is:
t = (v_f - v_i) / g
where:
t is the time taken
v_f is the final velocity (which is zero at the maximum height)
v_i is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the initial velocity is 18 m/s and the final velocity is 0 m/s (at maximum height). Substituting these values, we get:
t = (0 - 18) / (-9.8)
t = -18 / -9.8
t ≈ 1.84 seconds
This means it takes approximately 1.84 seconds for the ball to reach its maximum height.
To calculate the total time the ball is in the air, we need to consider both the upward and downward journeys. Since the upward journey takes 1.84 seconds, the downward journey will also take around 1.84 seconds (assuming no air resistance).
Therefore, the total time the ball is in the air will be:
Total time = Time up + Time down
Total time = 1.84 seconds + 1.84 seconds
Total time ≈ 3.68 seconds
Therefore, the ball will be in the air for less than 4 seconds when tossed upward at 18 m/s.