it would be..... C
sorry if I am wrong I tryed to think, At least I try!
Answer:0.58 m
Explanation:
The initial velocity of the ball is u = 2.0 m/s
The height of the table is, h = 1.0 m
The ball falls in vertical direction under acceleration due to gravity.
Time taken for ball to hit the floor:
h= ut + 0.5gt² ( from the equation of motion)
1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²
Solving this for t,
t = 0.29 s ( we have neglected the negative value of t)
In the same time, the ball would cover a horizontal distance of :
s = u t
⇒s = 2.0 m/s×0.29 s = 0.58 m
Thus, the landing spot is 0.58 m away from the table.
Explanation:
Below is an attachment containing the solution.
Answer:
2.72 km
Explanation:
(12.33 km)/ 1 hr * (1 hr)/ 60 min
0.2055 km/ min
distance=rate * time (assuming v is constant,
a=0)
(0.2055 km/ min)*(13.22 min)
2.72 km OR 2716.71 m
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
