A ball is thrown at an angle θ above the horizontal. In the presence of the acceleration due to gravity, how far below the straight-line path following its original direction of motion will the ball be at the end of 1.0 second?
A) 20 m
B) 4.9 m
C) 9.8 m
D) 2.5 m
E) None of the other choices is correct.

Question

Quizplus · Accepted Answer

The vertical displacement due to gravity after 1 second is given by the formula $ \frac{1}{2}gt^2 $, where $ g = 9.8 \, \text{m/s}^2 $. Substituting $ t = 1 \, \text{s} $, the displacement is $ \frac{1}{2} \times 9.8 \times 1^2 = 4.9 \, \text{m} $.

A Ball Is Thrown at an Angle θ Above the Horizontal