A ball dropped from a sixth floor window accelerates at 10 m/s^2
After 2 seconds it hits the
ground. It bounces back with a
velocity of −15 m/s
1.find the height from which the ball was dropped.
2. How high did it bounce?
h=gt²/2 = 10•2²/2=20 m
v=gt =10•2 =20 m/s
v=v₀-gt₁
v=0
t₁= v₀/g=15/10=1.5 s
h₁=v₀t₁-gt₁²/2 =
=15•1.5 -10•1.5²/2=11.25 m
a=Δv/Δt= (-15-20)/0.02= - 1750 m/s
Sure! Let's solve the problems step-by-step.
1. To find the height from which the ball was dropped, we can use the equation of motion:
h = ut + (1/2)at^2
Where:
h is the height from which the ball was dropped,
u is the initial velocity (which is 0 since the ball was dropped),
a is the acceleration (given as 10 m/s^2),
and t is the time taken for the ball to hit the ground (given as 2 seconds).
Plugging in the values, we get:
h = 0 * 2 + (1/2) * 10 * (2^2)
h = 0 + 0.5 * 10 * 4
h = 20 meters
Therefore, the height from which the ball was dropped is 20 meters.
2. To find how high the ball bounced, we can use the equation of motion again:
v^2 = u^2 + 2ah
Where:
v is the final velocity of the ball after bouncing (given as -15 m/s),
u is the initial velocity before bouncing (0 since the ball was dropped),
a is the acceleration due to gravity (which is -10 m/s^2 since the ball is moving upwards),
and h is the height the ball bounces.
We can rearrange the equation to solve for h:
h = (v^2 - u^2) / (2a)
Plugging in the values, we get:
h = (-15^2 - 0) / (2 * (-10))
h = (225 - 0) / (-20)
h = -225 / 20
h = -11.25 meters
However, we can't have a negative height, so the ball cannot bounce to a negative height. Therefore, the ball does not bounce, and the answer is 0 meters.
So, the height from which the ball was dropped is 20 meters, but it does not bounce.
To find the height from which the ball was dropped, we can use the kinematic equation for vertical motion:
h = h0 + v0t + (1/2)at^2
Where:
h = final height (which is zero when the ball hits the ground)
h0 = initial height (what we want to find)
v0 = initial velocity (equal to zero since the ball was dropped)
a = acceleration (which is -10 m/s^2, considering the direction of motion)
t = time taken (which is 2 seconds)
Using the equation, we can substitute the given values:
0 = h0 + 0 + (1/2)(-10)(2)^2
Simplifying this equation, we get:
0 = h0 - 20
Rearranging the equation, we find:
h0 = 20 meters
Therefore, the ball was dropped from a height of 20 meters.
To find how high the ball bounces, we can use the same equation, but this time with the given values:
h = h0 + v0t + (1/2)at^2
Here, we want to find the final height (h), so we substitute the given values:
h = 0 + (-15)(-2) + (1/2)(-10)(-2)^2
Simplifying:
h = 0 + 30 + 20
h = 50 meters
Therefore, the ball bounces to a height of 50 meters.