A ball is thrown straight up with an initial velocity of 50 ft/sec from the top of a two-story house that is 25 feet tall. The position equation for the ball is s(t)= -16t^2+50t+25

Question

(a) Find the velocity and acceleration of the ball after 1 second.

(b) Find when the ball will reach its maximum height.

(c) Find the maximum height of the ball.

(d) Find the time when the ball will hit the ground.

(e) Find the average velocity of the ball from the time it is thrown thill the time it hits the ground.

Answer 1

s(t)=-16t^2+50t+25

a) s(1)=-16+50+25=59

b) this is a parabola opening downward, so it has a maximum point, the vertex, (h,k), where h is the maximizing number and k is the maximum

h=-b/2a=-50/-32=1.5625 seconds

c) s(1.5625)=-16(1.5625)^2+50(1.5625)+25=-16(2.44140625)+78.125+50=

-39.06+78.125+25=142.19 feet rounded off

d) 0=-16t^2+50t+25, 16t^2-50t-25=0

use the quadratic formula to find t

e) you can try this one