Question

Small frogs that are good jumpers are capable of remarkable acceleration. One species of small frog reaches a takeoff speed of 3.7 m/s in 60 ms.
How high can the frog jump given it's acceleration is 62 m/s^2?

Answer 1

Hi there!

Begin by solving for the time it takes the frog to reach the top of its trajectory. (Vertical velocity = 0 m/s)

$\large\boxed{v_f = v_i + at}$

vf = final velocity (0 m/s)

vi = initial velocity (3.7 m/s)

a = acceleration due to gravity (9.8 m/s²)

t = time (s)

Plug in the known values:

$0 = 3.7 - 9.8t\\\\-3.7/(-9.8) = t = 0.378 s$

Now, we can use the following kinematic equation:

$\large\boxed{d = v_it + \frac{1}{2}at^2}$

d = displacement (m)

vi = initial velocity (3.7 m/s)

a = acceleration (9.8 m/s²)

t = time (0.378 sec)

Plug in the given values:

$d = 3.7(0.378) - \frac{1}{2}(9.8)(0.378)^2 = \boxed{0.698 \approx 0.7m}$