a ball is thrown upwards from a tower standing 300 feet tall. the initial velocity of the ball is 32 feet/second. the negative acceleration of the ball due to gravity is -16 in terms of feet per second squared. What is the velocity of the ball after 1 second?
h = -16t^2 + 32t + 300 , from your data
v = dh/dt
= -32t + 32
so when t = 1
v = -32 + 32 = 0
To find the velocity of the ball after 1 second, we can use the equation of motion for vertical motion under constant acceleration:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 32 feet/second
a = -16 feet/second squared (negative because acceleration due to gravity is downward)
t = 1 second
Substituting the given values into the equation:
v = 32 + (-16) * 1
v = 32 - 16
v = 16 feet/second
Therefore, the velocity of the ball after 1 second is 16 feet/second.