Gn n n n nnnhnhnhnhnhnhnhnhnhnhn
Answer:
time rising = 34 / 9.8 = 3.47 sec
total time in air = 2 * 3.47 sec = 6.94 sec
(time rising must equal time falling)
R = 17 m/s * 6.94 s = 118 m
Can also use range formula
R = v^2 sin (2 theta) / g
tan theta = 34 / 17 = 2
theta = 63.4 deg
2 theta = 126.9 deg
sin 126.9 = .8
v^2 = 17^2 + 34^2 = 1445 m^2/s^2
R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above
Answer:
Force, F = 900 N
Explanation:
Given that,
Mass of the football, m = 0.55 kg
Initial speed of the ball, u = 0
Final speed, v = 54 m/s
Time of contact, t = 0.033 s
We need to find the magnitude of the force exerted by the ball on the kicker’s foot. It is a case of second law of motion. It is equal to the rate of change of momentum. So,
So, the magnitude of the force exerted by the ball on the kicker’s foot is 900 N.
Answer:
1 m/s² down
Explanation:
"Calculate the acceleration of the elevator, and find the direction of acceleration."
Sum of forces in the +y direction:
∑F = ma
N − mg = ma
0.9 mg − mg = ma
-0.1 mg = ma
a = -0.1 g
a = -1 m/s²
We can first obtain time of flight from vertical fall
Initial velocity U=0, d = 6 m, a = 9.8 m/s²
d = ut + 1/2 at²
6.0 = 0 + (1/2 × 9.80 t²)
t = √(12/9.8)
= 1.106 sec
Horizontal velocity = Vh = Dh/t
= 24.0 /1.106 s
= 21.69 m/s
The ball was thrown at a speed of 21.69 m/s
