a) 70.1 m
The ball is moving by uniformly accelerated motion, with constant acceleration 
(acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation
where 
is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:
b) 3.9 s
We can still use the same equation we used in the previous part of the problem:
This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have
And solving for t we find
Answer:
r = Q/(8s^2)
Step-by-step explanation:
Divide by the coefficient of r.
Q/(8s^2) = 8rs^2/(8s^2) = r
Answer:
Thus, the ratio is:
a:b = 1:4
Step-by-step explanation:
Given the expression
Divide both sides by 2
Divide both sides by b
Thus, the ratio is:
a:b = 1:4
Step-by-step explanation:
I think all of you under stand this question so please like the answer
I think it is 31 I might be wrong but I tried :)
