Answer:
u = 104.68 m/s
Explanation:
given,
horizontal distance = 150 m
elevation of 12.4 m
angle = 8.6°
horizontal motion = x = u cos θ. t .............(1)
vertical motion =
................(2)
from equation(1) and (2)
..........{3}
u = 104.68 m/s
The initial speed of the ball is u = 104.68 m/s
Answer:
9.875
Explanation:
w=f×s
395=40×s
make s the subject of the formula
s=395/40
=9.875
I would use miles
Because miles will give you less of an answer
90+66=156
156/2=78
Reply:78kilometers in 2 hours.
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
