Question

A camper dives from the edge of a swimming pool at water level with a speed of 8.0 m/s at an angle of 30.0° above the horizontal. (a) How long is the diver in the air? (b) How high is the diver in the air? (c)How far out in the pool does the diver land?

Answer 1

(a) 0.816 s

(b) 0.816 m

(c) 5.656 m

Step-by-step explanation:

u = 8 m/s, theta = 30 degree,

(a) Use the formula for time of flight

T = 2 u Sin theta / g

T = ( 2 x 8 x Sin 30 ) / 9.8 = 0.816 s

(b) Use the formula for maximum height

H = u^2 Sin^2theta / 2 g

H = ( 8 x 8 x Sin^2 30) / ( 2 x 9.8)

H = 0.816 m

(c) Use the formula for horizontal range

R = u^2 Sin 2 theta / g

R = ( 8 x 8 x Sin 60) / 9.8

R = 5.656 m