x=-16 to the second power+63T4+4 after lounch how many seconds will it take to hit the ground

To calculate the time it takes for an object to hit the ground, we can use the formula for the time of flight of a projectile. The formula is:

t = sqrt((2 * h) / g),

where:
- t is the time of flight (in seconds),
- h is the initial vertical displacement or height (in this case, h = 0 because it hits the ground),
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

In the given equation, x = -16t^2 + 63t + 4, we see that the height (h) is 0 when the object hits the ground. So we need to solve for t when h = 0.

To solve the equation -16t^2 + 63t + 4 = 0, we can use the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a),

where:
- a, b, and c are coefficients of the quadratic equation in standard form.

In this case, a = -16, b = 63, and c = 4. Plugging these values into the quadratic formula:

t = (-63 ± sqrt(63^2 - 4(-16)(4))) / (2(-16)).

Simplifying:

t = (-63 ± sqrt(3969 + 256)) / (-32),
t = (-63 ± sqrt(4225)) / (-32),
t = (-63 ± 65) / (-32).

This gives us two solutions for t:

t1 = (-63 + 65) / (-32) = 2 / (-32) = -1/16,
t2 = (-63 - 65) / (-32) = -128 / (-32) = 4.

Since time cannot be negative in this context, the time it takes for the object to hit the ground is 4 seconds.