Question

PHYSICS, HELP PLZ??!! 100PTS

Answer 1

Hi there!

We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:

$t_{max} = \frac{vsin\theta}{g}$

Where:

tmax = (? sec)

vsinθ = vertical comp. of velocity = 10sin(48) = 7.43 m/s)

g = acceleration due to gravity (9.8 m/s²)

We can solve for this time:

$t_{max} = \frac{7.43}{9.8} = 0.758 s$

When the ball is at the TOP of its trajectory, its VERTICAL velocity is equivalent to 0 m/s. Thus, we can consider this a free-fall situation.

We must begin by solving for the maximum height reached by the ball using the equation:

$d = y_0 + v_{0y}t + \frac{1}{2}at^2$

d = displacement (m)

vi = initial velocity (7.43 m/s)

a = acceleration due to gravity

d = displacement (m)

y0 = initial VERTICAL displacement (28m)

Plug in the values:

$d = 28 + 7.43(0.758) + \frac{1}{2}(-9.8)(0.758^2) = 30.817 m$

Now, we can use the rearranged kinematic equation:

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(30.817)}{9.8}} = 2.51 s$

Add the two times together:

$0.758 + 2.51 = \boxed{3.266 s}$