You throw a ball straight up in the air. The instant after leaving your hand the ball’s speed is 30 m/s. Ignoring the effects of air resistance, predict how fast the ball is traveling 2 seconds later? Where is it at this time?
h= vi*time-1/2 g t^2 is how high
Vf=vi-gt is how fast it is still going.
10.38
To predict how fast the ball is traveling 2 seconds later, we need to consider the effect of gravity on the ball's motion. When the ball is thrown straight up, its speed decreases due to gravity acting opposite to its motion until it reaches its highest point. At this point, the ball comes to a momentary stop before accelerating back downwards due to gravity.
Since we know the initial speed of the ball when it left your hand was 30 m/s, we can use the principle of conservation of energy to find the speed of the ball after 2 seconds.
At the highest point, the ball momentarily stops, which means its final speed will be the same as its initial speed of 30 m/s, but in the opposite direction because it's moving downwards. Therefore, the ball will have a speed of -30 m/s (negative sign indicating the change in direction) after 2 seconds.
Next, to determine the ball's position after 2 seconds, we need to calculate its displacement. The formula to calculate displacement can be written as:
displacement (d) = initial velocity (v) * time (t) + (1/2) * acceleration (a) * (time)^2
Since our initial velocity is 30 m/s and we're ignoring air resistance, the acceleration due to gravity is a constant -9.8 m/s^2 (taking the negative sign because it opposes the motion of the ball). Plugging in the values, we have:
d = 30 m/s * 2 s + (1/2) * (-9.8 m/s^2) * (2 s)^2
d = 60 m + (1/2) * (-9.8 m/s^2) * 4 s^2
d = 60 m - 19.6 m/s^2 * 4 s^2
d = 60 m - 78.4 m
d = -18.4 m
The negative sign in the displacement indicates that the ball is 18.4 meters below its initial position after 2 seconds. Therefore, the ball is traveling downward with a speed of -30 m/s and is located 18.4 meters below the initial position at this time.