Question

a ball is thrown upward from a height of 5.3 m at an initial speed of 70 m/sec. acceleration resulting from gravity is $-9.8m/sec^{2}$. neglecting air resistance, solve for the velocity $v(t)$ and the height $h(t)$ of the ball $t$ seconds after it is thrown.
velocity $v(t) = \square$ meters/second
height $h(t) = \square$ meters

Explanation:

Step1: Find velocity from acceleration

Velocity is the integral of acceleration.
$$v(t) = \int a(t) dt = \int -9.8 dt = -9.8t + C$$
Use initial velocity $v(0)=70$ to solve for $C$:
$$70 = -9.8(0) + C \implies C=70$$
So $v(t) = -9.8t + 70$

Step2: Find height from velocity

Height is the integral of velocity.
$$h(t) = \int v(t) dt = \int (-9.8t + 70) dt = -4.9t^2 + 70t + D$$
Use initial height $h(0)=5.3$ to solve for $D$:
$$5.3 = -4.9(0)^2 + 70(0) + D \implies D=5.3$$
So $h(t) = -4.9t^2 + 70t + 5.3$

Answer:

Velocity $v(t) = -9.8t + 70$ meters/second
Height $h(t) = -4.9t^2 + 70t + 5.3$ meters