Answer:
When the time of fall is doubled, the height of fall will be quadrupled
Explanation:
Given;
height of fall, h = d m
time of fall, t = t s
initial velocity of the object, u = 0 m/d
The height of fall of the object is calculated from the kinematic equation below;
where;
g is acceleration due to gravity, which is constant
if the time of fall is doubled, the height of fall is calculated as;
Therefore, when the time of fall is doubled, the height of fall will be quadrupled
Answer:
Explanation:
Use the equation
where h(t) is the height after a certain amount of time goes by, v0t is the initial upwards velocity, and h0 is the initial height of the projectile. For us:
h(t) = 10
v0t = 80
h0 = 3 and filling in:
and get everything on one side to factor:
This factors to
t = .09 sec and 4.9 sec. Let's interpret this.
The time of .09 is when the ball reached 10 feet on the way up, and
the time of 4.9 is when the ball reached 10 feet on the way back down. That's the height we need, 4.9 seconds.