Question

5. A person on a trampoline jumps straight up. When the person is 3 m above the surface and
with upward velocity of 2.5 m/s, drops a ball. a) What is the speed of the ball at the time of
impact? b) How long before the ball impacts the surface of trampoline?

Answer 1

a)The speed of the ball at the time of impact will be 8.1 m/sec.

b)The time taken by the ball impacts the surface of trampoline will be 1.1 second.

What is speed?

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. Its SI unit is m/sec.

Given data;

g( gravitational acceleration) = 9.81 m/s2

u (initial velocity of ball) = 2.5 m/s

h₁(initial height of the ball) = 3.0 m

v (speed of the ball at the time of impact ) ?

The time taken by ball for upward journey = u / g

t₁ = u / g

t₁  = (2.5 m/s) / (9.81 m/s2) =

t₁  =0.2548 s

From second equation of motion,upward vertical distance covered by ball is;

$\rm h_2= \frac{1}{2} gt^2\\\\ h_2=\frac{1}{2} \times 9.81 \times (0.2548)^2\\\\ h_2=0.3184$

The total distance in the downward direction travelled;

$\rm h_3=h_1+h_2 \\\\ h_3=3.0+0.3184 \\\\ h_3=3.3184 \ m$

The time taken for the downward motion is;

$\rm t_2 = (\frac{2 \times h_3 }{g} )^\frac{1}{2}\\\\ t_2 = 0.825 \ m/sec$

The speed of the ball at the time of impact is found as;

$\rm V=gt_2 \\\\ V= 9.81 \times 0.8225 \\\\ V= 8.07 \ m/sec$

The total time of the flight of ball is;

$\rm t_3 = t_1+t_2 \\\\ t_3=0.2548+0.8225 \\\\ t_3=1.0774 \ sec$

Hence,the speed of the ball at the time of impact will be 8.1 m/sec and time taken by the ball impacts the surface of trampoline will be 1.1 second.

To learn more about the speed refer to the link;

https://brainly.com/question/7359669

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