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3 years ago

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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

ment, consider a ball thrown off the top of a building with a velocity v at an angle θ with respect to the horizontal. Let the building be 44.0 m tall, the initial horizontal velocity be 8.60 m/s, and the initial vertical velocity be 10.5 m/s. Choose your coordinates such that the positive y-axis is upward, and the x-axis is to the right, and the origin is at the point where the ball is released.

1 answer:

Aleks04 [339]3 years ago

3 0

Explanation:

It is given that,

Height of the building, h = 44 m

Initial horizontal velocity, $u_x=ucos\theta=8.6\ m/s$

Initial vertical velocity, $u_y=usin\theta=10.5\ m/s$

It can be assumed to find the maximum height of the projectile and time taken to reach the maximum height.

The formula for the maximum height is given by :

$H=\dfrac{(u\ sin\theta)^2}{2g}$

g is the acceleration due to gravity

$H=\dfrac{(10.5)^2}{2\times 9.8}=5.625\ m$

The total maximum height, h' = h + H

$h'=44+5.625=49.625\ m$

The formula for the time of flight is given by :

$t_{max}=\dfrac{u\ sin\theta}{g}$

$t_{max}=\dfrac{10.5}{9.8}=1.07\ s$

Hence, this is the required solution.

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