Can you help me with this questions please?

Question

Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation

​a(t)equals=v primev′​(t)equals=​g,

where

gequals=minus−9.8 m divided by s squared9.8 m/s2.

a. Find the velocity of the object for all relevant times.

b. Find the position of the object for all relevant times.

c. Find the time when the object reaches its highest point. What is the​ height?

d. Find the time when the object strikes the ground.

A softball is popped up vertically​ (from the​ ground) with a velocity of

26 m divided by s26 m/s.

Answer 1

a. Accelaration a = - g = - 9.8 M/s^2; then velocity v = ∫(-9.8)dt = - 9.8t + C.Now because at t= 0 v = 26 m/s we have C = 26 and

v = 26 - 9.8t m/s

b. Position y = ƒvdt = ∫(26 - 9.8t)dt = 26t - 4.9t² + C; at t = 0 y = 0 and so C = 0

so position y = 26t - 4.9t^2^.

c. At rhe top of moution v = 0 and 26 - 9.8t = 0, t = 26/9.8 = 2.653 s;

h_{max = y(2.653) = 26·2.653 - 4.9·2.653}² = 34.49 m

d. On the ground y = 0, or 26t - 4.9t² = 0 t = 0 (moment when object was thrown, t = 26/4.9 = 5.31 s