Answer :
Answer:
She must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.
Step-by-step explanation:
Given:
The angle of projection of the projectile is, [tex]\theta =65[/tex]°
Range of the projectile is, [tex]R=520[/tex] m.
Acceleration due to gravity, [tex]g=9.8\ m/s^2[/tex]
The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.
The range of projectile is given as:
[tex]R=\frac{v_{0}^2\sin2\theta}{g}[/tex]
Plug in all the given values and solve for minimum speed, [tex]v_0[/tex].
[tex]520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}[/tex]
Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.
The minimum speed with which Daring Darless must be launched to cross the 520-m gap is 81.6 m/s.
What minimum speed must she be launched to cross the 520-m gap?
We know that for any projectile motion the range is given by the formula,
[tex]R =\dfrac{u^2\ sin\ 2\theta}{g}[/tex]
Given to us
θ = 65°
Horizontal distance (Range), R = 520 m
g = 9.81 m/s²,
[tex]520 =\dfrac{u^2\ sin\ 2(65^o)}{9.81}\\\\u = 81.6\rm\ m/s[/tex]
Thus, the minimum speed with which Daring Darless must be launched to cross the 520-m gap is 81.6 m/s.
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