#### To determine

**To find:** The velocity after 2 s and after 4 s if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.

#### Answer

The velocity after 2 s is 4.9 m/s.

The velocity after 4 s is −14.7 m/s.

#### Explanation

The height of the particle is, h(t)=2+24.5t−4.9t2.

Find the derivative of h(t) and obtain the velocity.

v(t)=h'(t)=24.5−9.8t

Thus, the velocity is v(t)=24.5−9.8t (1)

Substitute *t* = 2 in equation (1) and obtain the velocity after 2 s.

v(2)=24.5−9.8(2)=24.5−19.6=4.9

Thus, the velocity after 2 s is 4.9 m/s.

Substitute *t* = 4 in equation (1) and obtain the velocity after 4 s.

v(4)=24.5−9.8(4)=24.5−39.2=−14.7

Thus, the velocity after 2 s is −14.7 m/s.

#### To determine

**To find:** when the projectile reach its maximum if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.

#### Answer

The projectile reaches its maximum height at 2.5 s.

#### Explanation

From part (a) the velocity of the particle is v(t)=24.5−9.8t.

Set v(t)=0 and obtain the t value.

24.5−9.8t=09.8t=24.5t=2.5

Thus, the projectile reaches its maximum height at *t* = 2.5 s.

#### To determine

**To find:** when the projectile reach its maximum if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.

#### Answer

The maximum height of the projectile is 32.625 m.

#### Explanation

From part (b), the projectile reaches its maximum height at t = 2.5 s.

Substitute t = 2.5 s for h(t) and obtain the maximum height of the projectile.

h(2.5)=2+24.5(2.5)−4.9(2.5)2=2+61.25−30.625=32.625

Thus, the maximum height of the projectile is 32.625 m.

#### To determine

**To find:** When the projectile hits the ground if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.

#### Answer

The projectile hits the ground approximately at *t* = 5.08 s.

#### Explanation

The projectile hits the ground at h(t)=0.

Set h(t)=0 and solve for *t*.

2+24.5t−4.9t2=04.9t2−24.5t−2=0

Use quadratic formula and solve further as follows.

t=24.5±(−24.5)2−4(4.9)(−2)2(4.9)=24.5±600.25+39.22(4.9)=−24.5±25.289.8≈5.08 or −0.07

Ignore the negative value, the value of *t* is approximately 5.08.

Thus, the projectile hits the ball when *t*= 5.08 s.

#### To determine

**To find:** With what velocity the projectile hits the ground if the height of the projectile shot vertically upward from 2 m above the ground level is h(t)=2+24.5t−4.9t2 with an initial velocity of 24.5 m/s.

#### Answer

The projectile hits the ground with the velocity −25.3 m/s.

#### Explanation

From part (d), the projectile hits the ground at h(5.08).

Substitute *t* = 5.08 for v(t) and obtain the required velocity.

v(5.08)=24.5−9.8(5.08)=24.5−49.784=−25.284

Thus, the projectile hits the ground with the velocity of −25.284 m/s.