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Selve the problem. 9) The function \( h \) described by \( h(t)=-16 t^{2}+33.1 t+124.26 \) gives the height of a ball thrown upward with a speed of 33.1 feet per second from a 124.26 ft high window \( t \) seconds after it is thrown until it hits the ground. Find the height of the ball 0.9 seconds after it is thrown.

Ask by Mckenzie Lee. in the United States
Feb 26,2025

Solución de inteligencia artificial de Upstudy

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The height of the ball 0.9 seconds after it is thrown is approximately 141.09 feet.

Solución

To find the height of the ball 0.9 seconds after it is thrown, substitute t = 0.9 into the function:   h(t) = –16t² + 33.1t + 124.26 Step 1: Calculate (0.9)²:   (0.9)² = 0.81 Step 2: Compute –16 times 0.81:   –16 × 0.81 = –12.96 Step 3: Compute 33.1 times 0.9:   33.1 × 0.9 = 29.79 Step 4: Add all the terms:   h(0.9) = –12.96 + 29.79 + 124.26       = (–12.96 + 29.79) + 124.26       = 16.83 + 124.26       = 141.09 Thus, the height of the ball 0.9 seconds after it is thrown is approximately 141.09 feet.

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The Deep Dive

To find the height of the ball 0.9 seconds after it is thrown, we need to substitute \( t = 0.9 \) into the function \( h(t) = -16t^2 + 33.1t + 124.26 \). Calculating step-by-step: 1. Calculate \( t^2 = 0.9^2 = 0.81 \). 2. Substitute \( t \) into the equation: \[ h(0.9) = -16(0.81) + 33.1(0.9) + 124.26 \] 3. Simplify each term: - \( -16 \cdot 0.81 = -12.96 \) - \( 33.1 \cdot 0.9 = 29.79 \) 4. Adding these together: \[ h(0.9) = -12.96 + 29.79 + 124.26 = 140.09 \] Thus, the height of the ball 0.9 seconds after it is thrown is approximately **140.09 feet**. Get ready to catch it as it descends!

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