A ball is thrown upward with an initial velocity of 8.52 m/s. How high will it go?

Answer 1

Solution:

Ball is thrown vertically upward upward V₀ = 8.52 m/s, The height it will reach is the maximum height y_max. At that point the final velocity v_f = 0. Starting from the equation of vertical motion

y = -1/2gt² + V_ot + y_o Here the initial height y₀ = 0 and the equation reduces to

y = -1/2gt² + V₀t , taking the derivative of y wrt to time gives you the final velocity

dy/dt = V_f = -gt + V_o knowing that V_f = 0, then 0 = -gt + V_o , V_o = gt, t = V₀/g = 8.52 m/s / 9.81m/s² = 0.87 s

y_max = -1/2(9.81m/s²)( 0.87s)² + 8.52 m/s(0.87s) = 3.7 m

Another way to solve this is by using the equation V_f² - V_o² = 2a x

her a = -g and x = h = height, then this can be written as V_f² - V_o² = 2(-g) h knowing that V_f = 0 it reduces to

-V_o² = -2gh , V_o² = 2gh , h = V_o² / 2g = (8.52 m/s)² / 2(9.81 m/s²) = 72.59 m²/s² / 19.62 m/s² = 3.7 m

Same answer as above