Two boys are throwing a baseball back and forth. The ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. If acceleration due to gravity is –16 ft/s2, how high above the ground is the ball 2 s after it is thrown?
h(t)=at^2+vt+h0 <-- formula
g = -32 ft/s^2
it is -32/2 which is -16
and
h(2) = (1/2)(-32) t^2 + 36 t + 4
h(2) = -16 (2)^2 + 36 (2) + 4
h(2) = -64 + 72 + 4
h(2) = 12 feet
25ft^2
To find the height of the ball 2 seconds after it is thrown, we can use the formula:
h(t) = at^2 + vt + h0
Given:
Initial height, h0 = 4 ft
Upward velocity, v = 36 ft/s
Acceleration due to gravity, a = -16 ft/s^2 (negative because it is acting downwards)
Substituting these values into the formula, we have:
h(t) = -16t^2 + 36t + 4
To find the height after 2 seconds, we can substitute t = 2 into the equation:
h(2) = -16(2)^2 + 36(2) + 4
h(2) = -16(4) + 36(2) + 4
h(2) = -64 + 72 + 4
h(2) = 12 + 4
h(2) = 16 ft
Therefore, the ball is 16 ft above the ground 2 seconds after it is thrown.
To find the height of the ball 2 seconds after it is thrown, we can use the formula for the height of an object in free fall:
h(t) = at^2 + vt + h0
Where:
h(t) is the height of the ball at time t
a is the acceleration due to gravity (-16 ft/s^2, since it acts downward)
t is the time in seconds
v is the initial upward velocity of the ball (36 ft/s, given)
h0 is the initial height of the ball (4 ft, given)
Plugging in the values into the formula:
h(2) = (-16)(2)^2 + (36)(2) + 4
Simplifying:
h(2) = -64 + 72 + 4
h(2) = 12 + 4
h(2) = 16 ft
Therefore, the ball is 16 feet above the ground after 2 seconds.