Question

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. After how long will it reach its maximum height? What is the maximum height?

A. After 4 seconds the object will reach its maximum height of 58.8 seconds
B. After 5 seconds the object will reach its maximum height of 38.3 seconds
C. After 6 seconds the object will reach its maximum height of 58.8 seconds
D. After 2 seconds the object will reach its maximum height of 78.4 s
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2. An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground?

A. -2 seconds
B. 2 seconds
C. 6 seconds
D. 78.4 seconds

Answer 1

Step-by-step D. After 2 seconds the object will reach its maximum height of 78.4 m

C. 6 seconds,

Step-by-step explanation:

s(t) = –4.9t² + 19.6t + 58.8

s(t) = –4.9t² + 19.6t + 58.8

4.9 x (t - 2)² = 0

t = 2 hopes this helps

Answer 2

D. After 2 seconds the object will reach its maximum height of 78.4 m

C. 6 seconds, the object strike the ground

Explanation:

s(t) = –4.9t² + 19.6t + 58.8

max s(t) = 58.8 - (19.6²/ 4* (-4.9)) = 58.8 + 19.6 = 78.4

78.4 = –4.9t² + 19.6t + 58.8

4.9t² - 19.6t + 19.6 = 0

4.9 x (t² - 4t + 4) = 0

4.9 x (t - 2)² = 0

t = 2

When it strike the ground, the height = 0

–4.9t² + 19.6t + 58.8 = 0

t² - 4t - 12 = 0

(t - 6) (t + 2) = 0 t should be positive

t = 6