A girl throws a tennis ball straight into the air with a velocity of 64 feet/sec. If acceleration due to gravity is -32 ft/sec2, how many seconds after it leaves the girl's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
4 secs
2 secs
1 sec
Cannot be determined
the height y is given by
y = -16t^2+64t
as with all y=ax^2+bx+c
the vertex (max in this case) is at -b/2a = 2
change in up velocity/time = -32 ft/s / s
original up velocity = 64
so when will the velocity up be zero???
v up = 64 - 32 t
0 = 64 - 32 t
t = 2 sconds
Thank you
To solve this problem, we need to understand the motion of the ball when it is thrown vertically upwards.
The initial velocity of the ball thrown vertically upwards is 64 ft/sec. The acceleration due to gravity is -32 ft/sec^2, which is negative because it acts opposite to the direction of motion.
At the highest point of its trajectory, the ball will momentarily come to rest before falling back down. This means that the final velocity at the highest point is 0 ft/sec.
We can use the equation of motion:
final velocity = initial velocity + (acceleration * time).
Substituting the given values, we have:
0 ft/sec = 64 ft/sec + (-32 ft/sec^2 * time).
Simplifying the equation, we have:
-64 ft/sec = -32 ft/sec^2 * time.
Dividing both sides by -32 ft/sec^2, we get:
2 sec = time.
Therefore, it will take the ball 2 seconds after leaving the girl's hand to reach its highest point.
So, the answer is 2 secs.