Please check your Eq for errors. The -49
should be -4.9.
should be -4.9.
Substituting the value of h in the equation, we have:
-49t^2 + 35t + 140 = 60
To solve this quadratic equation, we can rearrange it to the standard form:
-49t^2 + 35t + 80 = 0
Now, we can use the quadratic formula to find the value of t:
The quadratic formula is given by:
t = (-b ± sqrt(b^2 - 4ac))/(2a)
Here, a = -49, b = 35, and c = 80.
Substituting the values into the formula:
t = (-35 ± sqrt(35^2 - 4(-49)(80)))/(2(-49))
Simplifying further:
t = (-35 ± sqrt(1225 + 15680))/(-98)
t = (-35 ± sqrt(16905))/(-98)
Now, calculate the square root:
t = (-35 ± 130.2)/(-98)
We have two possible solutions:
1. t = (-35 + 130.2)/(-98)
t = 0.974 seconds (rounded to three decimal places)
2. t = (-35 - 130.2)/(-98)
t = 1.584 seconds (rounded to three decimal places)
Since we are looking for the time when the ball will be 60 meters from the ground, we only consider the positive time value, which is:
t = 0.974 seconds (rounded to the nearest tenth of a second)
Therefore, the ball will be 60 meters from the ground approximately 0.974 seconds after it was thrown.