a boy throws a ball from the top of a tower with velocity 20 m/s in horizontal direction. The speed of the ball after 2 seconds is?
V = Vo + g*t = 0 + 9.8*2 = 19.6 m/s.
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To find the speed of the ball after 2 seconds, we need to consider the horizontal and vertical components of the ball's motion separately.
First, let's look at the horizontal component. The ball is thrown horizontally, meaning there is no horizontal acceleration acting on it. Therefore, the horizontal component of the ball's velocity remains constant throughout its motion. In this case, the horizontal velocity is given as 20 m/s.
Next, let's consider the vertical component. The ball experiences a downward acceleration due to gravity, which is approximately 9.8 m/s^2 in most cases. Since the initial vertical velocity is zero (since the ball is only thrown horizontally), we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, u = 0 (since the initial vertical velocity is zero), a = 9.8 m/s^2 (acceleration due to gravity), and t = 2 s. Plugging in these values, we get:
v = 0 + 9.8 * 2
v = 0 + 19.6
v = 19.6 m/s
So, the speed of the ball after 2 seconds is 19.6 m/s.