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Chapter 2: Problem 27

A ball is projected vertically upwards such that it attains a height of \(h\) after \(5 \mathrm{~s}\) and \(9 \mathrm{~s}\) of its motion. The speed of projection is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (A) \(20 \mathrm{~ms}^{-1}\) (B) \(50 \mathrm{~ms}^{-1}\) (C) \(35 \mathrm{~ms}^{-1}\) (D) \(70 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The speed of projection is \(50 \mathrm{~ms}^{-1}\)

Step by step solution

01

Use the equation of motion

Start by applying the first equation of motion to find the initial velocity of the ball. The first equation of motion is \(v = u + g t\), where: \n- \(v\) is the final velocity, which we know is \(0 \, \mathrm{ms}^{-2}\), since at its peak, the object stops momentarily before starting to descend again, \n- \(u\) is the initial velocity, which we will designate as \(u\), \n- \(g\) is the acceleration due to gravity, which is \(-10 \, \mathrm{ms}^{-2}\) (negative because it's acting in the opposite direction of motion), \n- and \(t\) is the time, which is \(5s\) (time to reach the peak).
02

Substitute the values into the equation

Substitute the given numbers into the equation to solve for \(u\): \n\(0 = u - 10 \cdot 5 \) \nSolving this equation will give the initial velocity \(u\).
03

Solve for initial velocity

Solving the equation from step 2, we have: \n\( u = 10 \cdot 5 \),Thus, \(u = 50 \mathrm{~ms}^{-1}\)

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Most popular questions from this chapter

A boat travels from south bank to north bank of river with a maximum speed of \(8 \mathrm{~km} / \mathrm{h}\). A river current flows from west to east with a speed of \(4 \mathrm{~km} / \mathrm{h}\). To arrive at a point opposite to the point of start, the boat should start at an angle (A) \(\tan ^{-1}(1 / 2)\) west of north (B) \(\tan ^{-1}(1 / 2)\) north of west (C) \(30^{\circ}\) west of north (D) \(30^{\circ}\) north of west

A man holds an umbrella at \(30^{\circ}\) with the vertical to keep himself dry. He, then, runs at a speed of \(10 \mathrm{~ms}^{-1}\) and finds the rain drops to be hitting vertically. Speed of the rain drops with respect to the running man and with respect to earth are (A) \(20 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\) (B) \(10 \mathrm{~ms}^{-1}, 20 \sqrt{3} \mathrm{~ms}^{-1}\) (C) \(10 \sqrt{3} \mathrm{~ms}^{-1}, 20 \mathrm{~ms}^{-1}\) (D) \(20 \mathrm{~ms}^{-1}, 10 \sqrt{3} \mathrm{~ms}^{-1}\)

A boat, which has a speed of \(5 \mathrm{~km} / \mathrm{h}\) in still water, crosses a river of width \(1 \mathrm{~km}\) along the shortest possible path in 15 minutes. The velocity of the river water in kilometers per hour is (A) 1 (B) 3 (C) 4 (D) \(\sqrt{41}\)

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A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=\) \(8 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is (A) \(2.62 \mathrm{~m} / \mathrm{s}\) (B) \(4.6 \mathrm{~m} / \mathrm{s}\) (C) \(3.57 \mathrm{~m} / \mathrm{s}\) (D) \(1.414 \mathrm{~m} / \mathrm{s}\)
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