A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i
1 answer:
Answer:
The total time it is in the air for the ball is 1.6326 s
Given:
Initial velocity = 8
To find:
the total time it is in the air = ?
Formula used:
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
Solution:
A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.
The time taken by the ball to reach the maximum height is given by,
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
t =
t = 0.8163 s
Thus, time taken by the ball to reach the ground again = time taken to reach maximum height
So, Total time required for ball to reach ground = 2t = 2 × 0.8163
Total time required for ball to reach ground = 1.6326 s
The total time it is in the air for the ball is 1.6326 s
You might be interested in
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²