an object is thrown upwards at time t = 0. After t seconds, its height is y = - 4.9t^ 2 + 7t +1.5 meters above the ground

t = -7 / (2 * (-4.9)) t ≈ 0.714 Therefore, the maximum height occurs at t ≈ 0.714 seconds. To find the maximum height, we can substitute this value of t into the height function: y = -4.9(0.714)^2 + 7(0.714) + 1.5 y ≈ 4 meters Therefore, the maximum height of the object is approximately 4 meters above the ground.