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Question
Solve the following problem.
A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50 m? Assume g = 10 m s-2.
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Solution
Given: 40 ≤ R ≤ 50, θ = 30°, g = 10 m/s2
To find: Range of initial velocity (u)
Formula: R = `("u"^2 sin (2theta))/"g"`
Calculation: From formula,
The range of initial velocity,
`40 <= ("u"^2 sin (2theta))/"g" <= 50`
∴ `"40g"/(sin (2theta)) <= "u"^2 <= "50g"/(sin(2theta))`
∴ `sqrt(("40g")/(sin (2theta))) <= "u" <= sqrt(("50g")/(sin(2theta))`
∴ `sqrt((40 xx 10)/(sin (60))) <= "u" <= sqrt((50 xx 10)/(sin (60)))`
∴ 21.49 m/s ≤ u ≤ 24.03 m/s
The range of initial velocity should be between 21.49 m/s ≤ u ≤ 24.03 m/s.
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