Projectile Motion

Question

question 1: An object is thrown upward with an initial velocity of 15 feet per second. Its height is given by

h= -15t^2 + 60t. After how many seconds does it hit the ground?

question 2: A projectile is thrown upward so that its distance above the ground after t seconds is h= -15t^2 + 60t. What is its maximum height?

Answer 1

there's an ambiguity or contradiction in the problem

h(t) =-15t^2 +60t

take the derivative and set equal to zero, then solve for t

h'(t) =-30t + 60 = 0

30t=60

t = 2 seconds

plug that t value back into the h(t) equation

max height = h(2) = -15(2)^2 + 60(2) =-60+120= 60 feet

maximum height = 60 feet

h(t) = -15t^2 +60t = 0 for when it hits the ground

divide by -15 to get

t^2 -4t = 0

factor

t(t-4) = 0

set each factor = 0 and solve for t

t = 0 and 4 seconds. in 2 seconds it reaches maximum height, and in 4 seconds it comes back to zero height

The ambiguity comes in where you state initial velocity is 15 feet per second. That's probably a mistake and you meant that initial velocity is 60 feet per second and acceleration is -30 feet per second per second.

actually it should be -16t^2 with -32 feet per second per second. At least on earth at sea level.

someone typed 5 when it should have been 6, -15 when it should have been -16. take the derivative of -16t^2 and you get -32t which represents the deceleration due to gravity of 32 feet per second per second.

Answer 2

You can plot y = -15x² + 60x on Desmos and read off the answers

The first is the x intercept and the second is the y coordinate of the vertex.

or if you need to do it algebraically

to find the x intercept solve -15x² + 60x = 0, so x (-15x + 60) = 0, so one of the factors must be zero.

if y = ax² + bx + c, here a = -15, b = 60, c = 0, to find y coordinate of vertex set x = -b/2a and find y.