I throw a tennisball in the air with the velocity 8m/s. how far up can the ball come
+23254 The formula: y = -4.9t² + vt can be used:
y = height v = initial velocity t = time (in seconds)
-4.9 m/sec² reflects the acceleration due to gravity, - because it pulls downward
y = -4.9t² + 8t
At t = 0, y = 0 and the ball begins to rise.
When y = 0 again, the ball has fallen back down to earth.
If you can find the time when the ball again hits the ground, then at half that time, the ball will be at its maximum height. So we need to solve the equation:
0 = -4.9t² + 8t Factor this by factoriing out the t:
0 = t(-4.9t + 8)
So: t = 0 and -4.9t + 8 = 0
-4.9t = -8
t = -8/-4.9 which is approximately t = 1.633
Halfway between these two times, the ball will be at its maximum height ---> t = 0.816
When t = 0.816 ---> y = (-4.9)(0.816)² + 8(0.816) ≈ 3.3 m
+23254 The formula: y = -4.9t² + vt can be used:
y = height v = initial velocity t = time (in seconds)
-4.9 m/sec² reflects the acceleration due to gravity, - because it pulls downward
y = -4.9t² + 8t
At t = 0, y = 0 and the ball begins to rise.
When y = 0 again, the ball has fallen back down to earth.
If you can find the time when the ball again hits the ground, then at half that time, the ball will be at its maximum height. So we need to solve the equation:
0 = -4.9t² + 8t Factor this by factoriing out the t:
0 = t(-4.9t + 8)
So: t = 0 and -4.9t + 8 = 0
-4.9t = -8
t = -8/-4.9 which is approximately t = 1.633
Halfway between these two times, the ball will be at its maximum height ---> t = 0.816
When t = 0.816 ---> y = (-4.9)(0.816)² + 8(0.816) ≈ 3.3 m
