The height h, in feet, of a ball above the ground t seconds after being thrown upward with a velocity of 48 ft/s is given by . After how many seconds will the ball be 35ft above the ground?
height = time * 48ft/s
35ft = time * 48ft/s
35ft/(48ft/s) = time
time = 0.72917 seconds
To find the time at which the ball is 35ft above the ground, we need to find the value of t when the height, h, is equal to 35ft.
First, let's write down the given equation for the height of the ball:
h(t) = -16t^2 + vt + s
Where:
h(t) is the height of the ball at time t,
-16 represents the gravitational constant (acceleration due to gravity in ft/s^2),
t represents time in seconds,
v is the initial velocity of the ball (48 ft/s in this case), and
s is the initial height of the ball (which is not given in this question).
Since the ball is thrown upward, the initial velocity v is positive (48 ft/s).
Now, substitute the given values into the equation:
h(t) = -16t^2 + 48t + s
We want to find the value of t when the height, h, is 35ft. So, we can write it as:
35 = -16t^2 + 48t + s
At this point, we need more information to proceed. Specifically, we need the initial height, s, of the ball. Without knowing the initial height, we cannot find the exact time at which the ball will be 35ft above the ground.