Question

Solution

Correct option is

96 m/s We now find values of t for which s(t) = 256. So

160t – 16t2 = 256

⇒ 16(t2 – 10t + 16) = 0

⇒ (t – 2) (t – 8) = 0

⇒ t = 2, t = 8

So          v(2) = 160 – 32.2 = 96

v(8) = 160 – 256 = –96.

So the velocity on the way up in 96 m/s.

SIMILAR QUESTIONS

Q1

The slope of the tangent to the curve represented by x = t2 + 3t – 8 and y = 2t2 – 2t – 5 at the point M (2, 1) is

Q2

The coordinates of the point P on the curve y2 = 2x3, the tangent at which is perpendicular to the line 4x – 3y + 2 = 0, are given by

Q3

The coordinates of points P(xy) lying in the first quadrant on the ellipsex2/8 + y2/18 = 1 so that the area of the triangle formed by the tangent at Pand the coordinate axes is the smallest, are given by

Q4

The points(s) on the curve y3 + 3x2 = 12y where the tangent is vertical is(are)

Q5

The equation of the common tangent to the curves y2 = 8x and xy = –1 is

Q6

If ab > 0 then the minimum value of Q7

The curve y = ax3 + bx2 + cx + 8 touches x – axis at P(2, 0) and cuts they – axis at a point Q where its gradient is 3. The value of a, b, c are respectively

Q8

If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, is

Q9

The tangent to the curve At the point corresponding to is