Answer:
A.) 62.5 ft
B.) 3.58 seconds
C.) 8.58 seconds
Explanation:
A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s
To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.
V^2 = U^2 - 2gH
Since the ball is going up, g will be negative. And at maximum height, V = 0
Substitute all the parameters into the formula
0 = 35^2 - 2 × 9.8 × H
19.6H = 1225
H = 1225/19.6
H = 62.5 ft
(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion
H = Ut - 1/2gt^2
Substitutes all the parameters into the formula
62.5 = 35t - 1/2 × 9.8 × t^2
62.5 = 35t - 4.9t^2
4.9t^2 - 35t + 62.5 = 0
Let's use quadratic equations to find t
Divide all by 4.9
t^2 - 7.143t + 12.755 = 0
t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2
( t - 3.57)^2 = 0.000102
( t - 3.57 ) = +/-( 0.01 )
t = 3.57 + 0.01
t = 3.58 seconds
Ignore the negative one.
(C) the total time tAC needed for it to reach the ground at C from the instant it is released.
When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft
Using equation 2 of equations of motion again.
h = 1/2gt^2
122.5 = 1/2 × 9.8 × t^2
122.5 = 4.9t^2
t^2 = 122.5/4.9
t^2 = 25
t = 5
Total time = 5 + 3.58 = 8.58 seconds
