During av softball hang s battery hits a ball update fun a initial height was 3 feet, the height in feet, the softball is g iven by s(t)=-16+70t+3.
I think there might be a mistake in your provided equation for the height of the softball, s(t)=-16+70t+3. The equation seems to be missing the variable for time t.
Assuming the equation is actually s(t) = -16t^2 + 70t + 3, where s(t) represents the height of the softball at time t, we can calculate the updated height when the softball hits the ball.
To find the time when the softball hits the ball, we need to set s(t) equal to 0 (since the height is 0 when the ball hits the ground). So, by solving the equation -16t^2 + 70t + 3 = 0, we can find the values of t when the softball hits the ball.
The equation can be factored as (-4t - 1)(4t - 3) = 0, which gives us the two possible values for t: t = -1/4 and t = 3/4. Since time cannot be negative in this context, we can consider t = 3/4 as the valid solution.
Plugging t = 3/4 into the equation s(t), we can find the height of the softball at that time:
s(3/4) = -16(3/4)^2 + 70(3/4) + 3
= -9 + 52.5 + 3
= 46.5
Therefore, the height of the softball when it hits the ball is 46.5 feet.