Answer:
1) The height of the ball from 0 to 5 seconds are;
At t = 0 second, height = 2
At t = 1 second, height h = 22.1
At t = 2 seconds, height h = 32.4
At t = 3 seconds, height h = 32.9
At t = 4 seconds, height h = 23.6
At t = 5 seconds, height h = 4.5
2) The correct option is;
D. -16·t² + 25·t + 1
Step-by-step explanation:
1) The equation of motion of the ball is given as follows;
H(t) = -4.9·t² + 25·t + 2
The height of the ball from 0 to 5 seconds are;
H(0) = -4.9×(0)² + 25×(0) + 2 = 2
H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1
H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4
H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9
H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6
H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5
Therefore, we have;
The height of the ball are
At t = 0 second, height = 2
At t = 1 second, height h = 22.1
At t = 2 seconds, height h = 32.4
At t = 3 seconds, height h = 32.9
At t = 4 seconds, height h = 23.6
At t = 5 seconds, height h = 4.5
2) Given that the equation of the ball is that of a projectile motion, such as follows;
h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t
it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1
.