Solve the problem first using the variables given and then the numerical quantit

Question

Solve the problem first using the variables given and then the numerical quantities. A rock is thrown up with an initial speed of vo= 26 m/s. The object hits the ground tp= 7.5 seconds later.

A) How much higher or lower is the launch point relative to the point where the rock hits the ground

B) Find the maximum height reached by the rock (relative to its bottom most point in the trajectory)

C) What is the velocity when it hits the ground?

D) Draw the acceleration vs time, velocity vs time and position vs time graphs

Explanation / Answer

given,

v0 = 26 m/s

time = 7.5 sec

let height of the launching point be x so

time to reach maximum height = 26 / 9.8

time to reach maximum height = 2.653 sec

maximum height relative to launcing point = 26^2 / (2 * 9.8)

maximum height relative to launcing point = 34.489 m

total distance = x + 34.489

time taken to cover this distance = sqrt(2(x + 34.489) / 9.8)

tota time = 7.5 so,

7.5 = sqrt(2(x + 34.489) / 9.8) + 2.653

x = 80.626 m

relative to ground height of launch point = 80.626 m

maximum height = 80.626 + 34.489

maximum height reached by the rock = 115.115 m

velocity = sqrt(2 * 9.8 * 115.115)

velocity when it hits the ground = 47.5 m/s

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