Question

Mr. Richard's science class is conducting a variety of experiments with projectiles and falling objects. Two groups of students are using a slingshot to launch tennis balls from ground level, and two groups are dropping tennis balls from the fire escape.
Below are descriptions of the parameters with which each group launches or drops their tennis balls.

Group A launches a tennis ball straight up with an initial velocity of 19 meters per second.
Group B launches a tennis ball straight up with an initial velocity of 50 feet per second.
Group C drops a tennis ball from a height of 19 meters.
Group D drops a tennis ball from a height of 50 feet.
Determine which group's activities can be modeled by the following equations, where h is the current altitude of the tennis ball.

h=-4.9t^2+19 Answer Options: Group A, Group B, Group C, Group D
h=-16t^2+50t
h=-16t^2+50
h=-4.9t^2+19t

Answer 1

In general, the height of a projectile is given by

$h= -\frac 1 2 g t^2 + v t + h_0$

where $h$ is the height at time $t$ of a projectile shot upward at velocity $v$ from initial height $h_0$.

The acceleration of gravity $g= 9.8 \textrm {m/s}^2 \textrm{ or } g=32\textrm{ feet/s}^2.$  We give it a negative sign in the equation because it pulls down.

Let's apply it to each group.

Group A launches a tennis ball straight up with an initial velocity of 19 meters per second.

We have MKS units, positive (upward) velocity and an initial height of 0.

$h= -\frac 1 2(9.8) t^2 + (19) t + 0 = -4.9t^2 + 19t$

Group B launches a tennis ball straight up with an initial velocity of 50 feet per second.    Not metric, velocity +50, initial height zero.

$h= -16t^2 + 50t$

Group C drops a tennis ball from a height of 19 meters.

MKS, initial height 19, initial velocity zero.

$h= -\frac 1 2 g t^2 + v t + h_0 = -4.9 t^2 + 19$

Group D drops a tennis ball from a height of 50 feet. 

$h= -16 t^2 + 50$

Answer 2

Answer:

C, B, D, A

Step-by-step explanation: