Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball be 2.0 second later?

78.4 m

Explanation:

Using newton's equation of motion,

S = ut + 1/2gt² ... Equation 1

Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.

Note: Taking upward to be negative, and down ward positive

Given: u = 49 m/s, t = 2.0 s, g = - 9.8 m/s²

Substitute into equation 1

S = 49 (2) - 1/2 (9.8) (2) ²

S = 98 - 19.6

S = 78.4 m

Hence the height of the ball two seconds later = 78.4 m