Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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]

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.