Question

A paperweight is thrown straight up with an initial upward velocity of 60 meters per second, and an initial height of 6 meters. The acceleration (due to gravity) of the object is given by a(t) = -9.8 meters per second per second, where tis in seconds. The maximum height of the paperweight is: =- O

A. 3.06 meters
B. 6.122 meters
C. 144.692 meters
D. -4.97 +60t+ 6 meters
E. 189.673 meters

Answer 1

The maximum height reached by the paperweight is found by using the kinematic equation H = V² / (2g) + initial height. After calculating with the given values, the maximum height is 189.673 meters, which corresponds to option E.

Step-by-step explanation:

The question is asking for the maximum height reached by a paperweight that is thrown straight up with a given initial velocity and height, considering the acceleration due to gravity. To find the maximum height, we need to use the kinematic equation that relates the initial velocity, acceleration, and the maximum height of a projectile, which is given by:

H = V² / (2g) + initial height,

where H is the maximum height, V is the initial upward velocity, and g is the acceleration due to gravity. Plugging in the values:

H = (60 m/s)² / (2 * 9.8 m/s²) + 6 m,

H = (3600 m²/s²) / (19.6 m/s²) + 6 m,

H = 183.673 m + 6 m,

H = 189.673 meters.

Therefore, the maximum height reached by the paperweight is 189.673 meters, which corresponds to option E.