If v(t) and h(t) denote the ball's velocity and height at time t, respectively, then we have h(0) = 1.2 m and v(0) = 58 m/s.
The ball is in free fall after being tossed, so its acceleration is constant at any time t :
[tex]a(t) = -9.8 \dfrac{\rm m}{\mathrm s^2}[/tex]
By the fundamental theorem of calculus, the velocity function is given by
[tex]\displaystyle v(t) = v(0) + \int_0^t a(u) \, du[/tex]
Compute the integral:
[tex]\displaystyle v(t) = 58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) \int_0^t du[/tex]
[tex]\displaystyle v(t) = 58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) (t - 0)[/tex]
[tex]\displaystyle \boxed{v(t) = 58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) t}[/tex]
Similarly, the height function is given by
[tex]\displaystyle h(t) = h(0) + \int_0^t v(u) \, du[/tex]
[tex]\displaystyle h(t) = 1.2 \, \mathrm m + \int_0^t \left(58 \frac{\rm m}{\rm s} - \left(9.8 \frac{\rm m}{\mathrm s^2}\right) t\right) \, du[/tex]
[tex]\displaystyle h(t) = 1.2 \, \mathrm m + \left(58 \frac{\rm m}{\rm s}\right) (t - 0) - \frac12 \left(9.8 \frac{\rm m}{\mathrm s^2}\right) (t^2 - 0^2)[/tex]
[tex]\displaystyle \boxed{h(t) = 1.2 \, \mathrm m + \left(58 \frac{\rm m}{\rm s}\right) t - \frac12 \left(9.8 \frac{\rm m}{\mathrm s^2}\right) t^2}[/tex]
Answer:
V(t) = 58-9.8t
h(t)=1.2+58t-4.9[tex]t^{2}[/tex]
Step-by-step explanation:
Using the standard formula
a=V f-Vi/t --rearrange
at=V f-Vi
V f=V i + a t
therefore
V(t)=58-9.8t
same with height
H= Hi+Vit+1/2at^2
therefore
h(t)=1.2+58t-4.9t^2
