How high would a ball go if it is thrown vertically upward and returns to launching position after 8 seconds?*1 pointA. 7.84 mB. 31.36 mC..78.4 mD. 313.6 m

To solve this problem, we need to use the physics formula for the height of an object under the influence of gravity, which is given by:

h = vt - 0.5g*t^2

where:

• h is the height,
• v is the initial velocity,
• t is the time, and
• g is the acceleration due to gravity.

Given that the ball returns to its launching position after 8 seconds, we know that the total time for the journey up and down is 8 seconds. Therefore, the time it takes for the ball to reach its highest point is half of this, which is 4 seconds.

We also know that the acceleration due to gravity is approximately 9.8 m/s^2.

Substituting these values into the formula, we get:

h = v4 - 0.59.8*4^2

We also know that at the highest point, the velocity of the ball is 0 (because it stops for an instant before falling back down). Therefore, we can set v = 0 and simplify the equation to:

h = 0 - 0.59.84^2 h = - 0.59.816 h = - 78.4

However, height cannot be negative, so we take the absolute value to get:

h = 78.4 m

Therefore, the ball would go 78.4 meters high. So, the correct answer is C. 78.4 m.