Answer:
a) 20s
b) 500m
Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.
To find time, we apply the UARM formula:
v final = (a x t) + v initial
Replacing the values gives us:
0 = (-10 x t) + 100
-100 = -10t
t = 10s
It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur
So 10s going up and another 10s going down:
10x2 = 20s
b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:
Δy = (1/2)(a)(t^2) + (v initial)(t)
Replacing the values gives us:
Δy = (1/2)(-10)(10^2) + (100)(10)
= (-5)(100) + 1000
= -500 + 1000
= 500 m
Hope this helps, brainliest would be appreciated :)
3.4m/s
Explanation:
Given parameters:
Distance to school = 14.4km
Time taken by Amy = 49min
Time taken by bill = 20min after Amy = 20+49 = 69min
Unknown parameters:
How much faster is Amy's average speed = ?
Solution:
Average speed is the rate of change of total distance with total time taken.
Average speed =
convert units to meters and seconds
1000m = 1km
60s = 1min
Distance to school = 14.4 x 1000 = 14400m
Time taken by Amy = 49 x 60 = 2940s
Time taken by Bill = 69 x 60 = 4140s
Average speed of Amy = = 4.9m/s
Average speed of Bill = = 1.4m/s
Differences in speed = 4.9 - 1.5 = 3.4m/s
Amy was 3.4m/s faster than Bill
learn more:
Average speed brainly.com/question/8893949
#learnwithBrainly
The force of earth's gravitational field is always directed downwards (towards the center of the earth. When the ball is thrown up, it is going against the earth's gravitational field and so, the earth's gravitational force pulls it back down, accelerating it downwards.
<span>answer under the link: http: //briskrange.com/7gAl
</span>