For t1:
t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec
For t2:
t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec
Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)
d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m
Where:
d = hor. distance
ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25
The answer is 3.19 : 2.25
So i believe is exercise:)
Answer:
a = 1.16 m/s²
Explanation:
In order to find the acceleration of the ball we will use 3rd equation of motion.
2as = Vf² - Vi²
where,
a = acceleration = ?
s = displacement = 21.9 m
Vf = Final Velocity = 7.14 m/s
Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)
Therefore, using the values, we get:
2a(21.9 m) = (7.14 m/s)² - (0 m/s)²
a = (50.97 m²/s²)/(43.8 m)
<u>a = 1.16 m/s²</u>
