Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

A 70.7 kg high jumper leaves the ground with
a vertical velocity of 6.7 m/s.
How high can he jump? The acceleration
of gravity is 9.8 m/s^2.
Answer in units of m

Sagot :

leena

Hi there!

We can use the following kinematic equation:

[tex]v_f^2 = v_i^2 + 2ad[/tex]

vi = initial velocity (6.7 m/s)
vf = final velocity (0 m/s at top of trajectory)

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

d = max height reached (? m)

We can plug in the knowns:

[tex]0 = 6.7^2 + 2(-9.8)(d)\\\\[/tex]

Solve for 'd':

[tex]d = \frac{6.7^2}{2(9.8)} = \boxed{2.29 m}[/tex]