Question

The height h, of an object that is thrown straight up at time t, is given by the following function: h(t) = d + ut - 1/2at^2, where: h(t) the height in metres, t = time in seconds, d = the initial height above the ground from which the object is thrown, u = the initial velocity in metres per second, and a = the acceleration of the object due to the force of gravity. Assume that the acceleration due to the force of gravity is 10 m/s ^2

Answer 1

An object is projected straight up from a platform situated 20 m above the ground at a speed of 50 m/s. Assuming the object reaches the ground, how many seconds after it projected does it reach the ground?

Answer 2

hint : when it reaches the ground, the height becomes 0

Answer 3

h(t) = d + ut - 1/2at^2

Answer 4

0 = d + ut - 1/2at^2

Answer 5

substitute given values and solve for t

Answer 6

Thank you so much. :)

Answer 7

yw