11. a ball is thrown upwards from a height of 2m with an initial velocity of 15m/s.\n(a) solve the…

Explanation:

Step1: Integrate (h''(t)) to find (h'(t))

Integrate (h''(t)=-9.8) with respect to (t). Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (h'(t)=\int h''(t)dt=\int-9.8dt=-9.8t + C_1). Given the initial condition (h'(0) = 15) (initial velocity), substitute (t = 0) and (h'(0)=15) into (h'(t)): (h'(0)=-9.8\times0 + C_1=15), so (C_1 = 15). Then (h'(t)=-9.8t + 15).

Step2: Integrate (h'(t)) to find (h(t))

Integrate (h'(t)=-9.8t + 15) with respect to (t). (h(t)=\int(-9.8t + 15)dt=-9.8\times\frac{t^{2}}{2}+15t+C_2=-4.9t^{2}+15t + C_2). Given the initial condition (h(0) = 2) (initial height), substitute (t = 0) and (h(0)=2) into (h(t)): (h(0)=-4.9\times0^{2}+15\times0 + C_2=2), so (C_2 = 2). Then (h(t)=-4.9t^{2}+15t + 2).

Step3: Solve for (t) when (h(t)=0) (part b)

Set (h(t)=-4.9t^{2}+15t + 2 = 0). Use the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (ax^{2}+bx + c = 0). Here (a=-4.9), (b = 15), (c = 2). (t=\frac{-15\pm\sqrt{15^{2}-4\times(-4.9)\times2}}{2\times(-4.9)}=\frac{-15\pm\sqrt{225 + 39.2}}{-9.8}=\frac{-15\pm\sqrt{264.2}}{-9.8}). We take the positive root (t=\frac{-15+\sqrt{264.2}}{-9.8}\approx\frac{-15 + 16.25}{-9.8}) (rejected) or (t=\frac{-15-\sqrt{264.2}}{-9.8}\approx\frac{-15-16.25}{-9.8}\approx3.19) s.

Step4: Find the time for maximum height (part c)

The velocity at maximum - height is (h'(t)=0). Set (h'(t)=-9.8t + 15=0). Solve for (t): (t=\frac{15}{9.8}\approx1.53) s. Substitute (t=\frac{15}{9.8}) into (h(t)): (h(\frac{15}{9.8})=-4.9\times(\frac{15}{9.8})^{2}+15\times\frac{15}{9.8}+2) (=-4.9\times\frac{225}{96.04}+ \frac{225}{9.8}+2) (=-\frac{1102.5}{96.04}+\frac{225}{9.8}+2) (=- 11.48+22.96+2=13.48) m.

Answer:

(a) (h'(t)=-9.8t + 15), (h(t)=-4.9t^{2}+15t + 2); (b) (t\approx3.19) s; (c) (t\approx1.53) s, (h\approx13.48) m.