Question

1. a ball is thrown upwards from a height of 2m with an initial velocity of 15m/s.

(a) solve the differential equation ( h(t)=-9.8m/s^{2} ), to find the velocity ( h(t) ) and height ( h(t) ) of the ball after ( t ) seconds.
(b) at what time ( t ) will the ball hit the ground?
(c) at what time ( t ) will the ball reach its maximum height? what is this maximum height?

Explanation:

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

Integrate \(h''(t)=-9.8\) with respect to \(t\).
Using the power rule \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), 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.