Dock diving is a great form of athletic competition for dogs of all shapes and sizes. Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 8.62 m/s at an angle of 28° with respect to the surface of the water. (Assume that the +x axis is in the direction of the run and the +y axis is up.) (a) If Sheba begins at a height of 0.85 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? m

3.139 m

Explanation:

Velocity = 8.62 m/s

Angle of flight = 28°

Height above water = 0.85 m

we are to take the +ve x-axis in the direction of run, and +ve y-axis up.

This is a projectile situation, and the horizontal distance of travel (range) is calculated as

R = (sin2∅)/2g

where g is the acceleration due to gravity = +9.81 m/ sine it acts downwards

R = (sin (2x28°))/(2 x 9.81)

R = (74.30 x 0.8290)/19.62

R = 61.60/19.62 = 3.139 m

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s