A ball is thrown upward from the top of a building which is 96ft tall w/ Initial velocity of 80 ft per sec.
Distance s(ft) of ball from the ground after t sec is s=96+80t-16t^(2)
1) how many sec. will the ball pass the top of the building when it is coming down?
How would I solve this?
set s(t) to 96 and solve.
96=96+80t-16t^2
solve for t
why would I set 96 for s(t) because that's the height of the building ?
but it's when the ball passes the top of the building
Also I tried solving it I got to
-16t(t-5)= 0 what should I do next
I finished solving it. thank you
riddle: my ones number is double my tens number, I'am less than 70 and more than 60
A ball is thrown vertically upward with velocity 100ft/sec and height of building is 40ft what is maximum height reached by ball and what is velocity when ball hits the ground
To solve this problem, we need to find the time at which the ball reaches the top of the building while coming down. This can be done by setting the equation for the height of the ball equal to the height of the building and solving for time.
Given:
Height of the building (h) = 96 ft
Initial velocity (u) = 80 ft/s
Acceleration due to gravity (g) = -32 ft/s^2 (since the ball is moving against gravity)
Let's start by setting up the equation for the height of the ball as a function of time (t):
s(t) = h + ut + (1/2)gt^2
Substituting the given values:
96 + 80t - 16t^2 = 96
Now, we will rearrange the equation to solve for time (t):
16t^2 - 80t = 0
Factoring out a common factor of 16t:
16t(t - 5) = 0
This equation is satisfied when either 16t = 0 or (t - 5) = 0.
From the equation 16t = 0, we find that t = 0. However, this does not provide a meaningful solution in this context.
From the equation (t - 5) = 0, we find that t = 5.
Therefore, the ball will pass the top of the building when it is coming down after 5 seconds.
Explanation:
To solve this problem, we used the equation for the height of the ball as a function of time, which is derived from the kinematic equation for motion under constant acceleration. By setting this equation equal to the height of the building, we found the time at which the ball reaches the top of the building while coming down. Finally, we solved the resulting quadratic equation to find the value of time.