Question 1a, 2b An object is launched at 19.6 meters per second (m/s) from a 58.

Question

Question 1a, 2b

An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation the object's height s at time t seconds after launch is s(t) = -4.9t^2 + 19.6t + 58.8, where s is in meters. a. To the nearest meter, what is the objects maximum height? b. When does the object strike the ground? An object is launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. Use general equation h(t) = -16t^2 + nu_0 t + h_0, where h(t) is height in feet, t is time in seconds, v_o is velocity, and h_0 is initial height. a. What will be the object's maximum height? b. When will it attain this height?

Explanation / Answer

a. The trajectory of the object given by the equation s(t) = -4.9 t2 +19.6t +58.8 is a parabola. The maximum height will be at the vertex of the parabola. For determining the vertex, we will convert the equation into vertex form. Wew have s(t) = -4.9 t2 +19.6t +58.8 = -4.9( t2 – 19.6t/4.9 – 58.8/4.9) = -4.9( t2 –4t -12) = -4.9( t2 –2*2t + 4 -16) = -4.9( t2 –2*2t + 4)+ 16*4.9 = -4.9(t-2)2 + 78.4. Thus, the vertex of the parabola is ( 2, 78.4). Therefore, the maximum height of the object (from the ground) is 78.4 meters.

1.b. The object will strike the ground when s(t) = 0 i.e. when -4.9 t2 +19.6t +58.8 or, on dividing both the sides by -4.9, when t2 – 4t – 12 = 0 or, when t2 -6t+2t-12= 0 or when t(t-6) +2(t-6) = 0 or when (t-6)(t+2) = 0. Since t cannot be negative, the object will hit the ground when t = 6 seconds i.e. after 6 seconds.

2. a. We have h(t) = -16t2 + v0 t +h0 = -16t2 + 64t +80 = -16(t2 –4t-5) = -16(t2 –4t +4 -9) = -16(t2 –4t +4) +144 = -16(t-2)2+144. This is the vertex form of the equation of a parabola. The vertex of the parabola is (2,144). Hence, the obect’s maximum height will be 144 meters.

b. When h(t) = 144, we have 144 = -16t2 + 64t +80 or, 16t2 -64 t +144-80 = 0 or, 16t2 -64 t + +64 =0 , or, on dividing both the sides by 16, we get t2 -4t +4 = 0 or, (t-2)2 = 0. Thus, the object will achieve maximum height when t = 2 seconds, i.e. after 2 seconds.

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