## University Calculus: Early Transcendentals (3rd Edition)

$$g_s=\frac{3}{4}(m/sec^2)$$
$$s(t)=15t-\frac{1}{2}g_st^2$$ - Find $v(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}\Big(15t-\frac{1}{2}g_st^2\Big)=15-g_st$$ The ball will reach its maximum height when its velocity decreases to $0$, which means $$v(t)=0$$ $$15-g_st=0$$ $$g_st=15$$ But we are already given that the ball reached its maximum height $20$ sec after being launched, so $t=20$: $$20g_s=15$$ $$g_s=\frac{3}{4}(m/sec^2)$$