An object is thrown into the air. The equation for the object's heights at time t seconds after launch is H(t) = -4.9t2 + 13.4t + 24.9, where H is in

meters.

Missing Part of the Question:

- The time required for the rocket to reach its maximum height

- The maximum height of the rocket

a. Time to reach maximum height is 1.37 seconds

b. Maximum Height = 34.1 metres

Step-by-step explanation:

Given.

H(t) = -4.9t² + 13.4t + 24.9

Solving (a): Time taken to reach maximum height.

A quadratic equation is:

f(x) = ax² + bx + c

Where

a = -4.9

b = 13.4

c = 24.9

This required time is represented by the vertex and is calculated as thus:

t = -b/2a

Substitute values for b and a

t = (-13.4)/(2 * -4.9)

t = -13.4/(-9.8)

t = 13.4/9.8

t = 1.37 (approximated)

Hence the time to reach maximum height is 1.37 seconds

b. Maximum height

Here, we simply substitute 1.37 for t in

H(t) = -4.9t²+ 13.4t + 24.9

Maximum Height = -4.9 * 1.37² + 13.4 * 1.37 + 24.9

Maximum Height = -4.9 * 1.8769 + 13.4 * 1.37 + 24.9

Maximum Height = -9.19681 + 18.358 + 24.9

Maximum Height = 34.06119

Maximum Height = 34.1 metres (approximated)

4x and 6y

im not sure though