The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Explanation:
Given:
v₀ = 22 m/s
v = 0 m/s
t = 17.32 s
Find: a
v = at + v₀
(0 m/s) = a (17.32 s) + (22 m/s)
a = -1.270 m/s²
Round as needed.
Answer:
No. Because it would correspond to zero Instantaneous acceleration.
Explanation:
hope this helps
Answer:
Decreases to half.
Explanation:
From the question given above, the following data were obtained:
Initial mass (m₁) = m
Initial force (F₁) = F
Initial acceleration (a₁) =?
Final mass (m₂) = ½m
Final force (F₂) = ¼F
Final acceleration (a₂) =?
Next, we shall determine a₁. This can be obtained as follow:
F₁ = m₁a₁
F = ma₁
Divide both side by m
a₁ = F / m
Next, we shall determine a₂.
F₂ = m₂a₂
¼F = ½ma₂
2F = 4ma₂
Divide both side by 4m
a₂ = 2F / 4m
a₂ = F / 2m
Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:
a₁ = F / m
a₂ = F / 2m
a₂ : a₁ = a₂ / a₁
a₂ / a₁ = F/2m ÷ F/m
a₂ / a₁ = F/2m × m/F
a₂ / a₁ = ½
Cross multiply
a₂ = ½a₁
From the illustrations made above, the acceleration of the car will decrease to half the original acceleration
Answer:
time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes
Explanation:
Given:
Time is inversely proportional to the speed
mathematically,
t ∝ (1/r)
let the proportionality constant be 'k'
thus,
t = k/r
therefore, for case 1
time = 3 hr
speed = 14 km/hr
3 = k/14
also,
for case 2
let the time be = t
r = 23 km/h
thus,
we have
t = k/23
on dividing equation 2 by 1
we get

or

or
t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)